Page 1 : DIFFERENTIAL EQUATIONS, , 3, , “He who seeks for methods without having a definite problem, , in mind seeks for the most part in vain”, D. HILBERT., , 2d 2 y  dy ,   0, dx 2  dx , , Example:, , Note: 1), dy, d2y, dny, y    y1 y   2  y 2 ............. y  n   n  y n, dx, dx, dx, 2) a differential equation involving more than one, independent variable is called a partial differential, equation, 2z, Example:,   xy  k, xy, ORDER OF A DIFFERENTIAL EQUATION, , Order of a differential equation is defined as the order, of the highest order derivative involved in the given, differential equation, 3, , HENRI POINCARE(1854-912), , 2, d3y, 2d y, , x,  0  ODE  3, , 2 , dx3,  dx , , Example:, , SAMPLE BLUE PRINT OF THIS CHAPTER, , GIR, OLA, PART PART PART PART PART, OFGIROLAMO, A, B, C, D, E, MO CARDANO NO., (16, CARDANO, (1h, CHAPTER-9, TEACHING, century), HOURS, DIFFERENTIAL, EQUATIONS, , 9, , 1M, , 2M, , 3M, , 5M, , -, , 1Q, , 1Q, , 1Q, , DEGREE OF A DIFFERENTIAL EQUATION, T.M, , 6M 4M, -, , -, , 10, , Degree of a differential equation is defined as the, highest power of the highest order derivative involved, in a differential equation when it is a polynomial, equation is called the degree of a differential equation, 3, , 3, 2, Example: d y  x 2  d y   0  deg ree  1, 3, 2, ,  dx , , dx, , CONTENTS, 1. Introduction, 2. Basic Concepts, 3. General and Particular Solutions of a, Differential Equation, 4. Formation of a Differential Equation whose, General Solution is given, 5. Methods of Solving First order, First Degree, Differential Equations, , Note: order and degree of a differential equation are, always positive integers, SOLVED PROBLEMS, TWO MARKS QUESTIONS, , 1. Find the order and degree of the differential, 2, , 3, , 3, 2, equations if defined  d y    d y    dy   y  0, 3, 2, ,  dx , ,  dx , ,  dx , , [Supp-14] [Pu Board Q.B-2m], TOPIC-1-ORDER AND DEGREE OF A DIFFERENTIAL, EQUATIONS, , Solution:, , 2. Find the order and degree of the differential, 3, 2, equations if defined  d y   2  d y    dy   0, 3, 2, , DIFFERENTIAL EQUATION, , An equation involving derivatives of the dependent, variable w.r.t independent variable is called a, differential equation, 2 xdy, Example:,  3y  5, dx, ORDINARY DIFFERENTIAL EQUATION, , A differential equation involving derivatives of the, dependent variable w.r.t only one variable is called an, ordinary differential equation, CHETHAN M G .SAI PU COLLEGE SHIMOGGA, , order-3 and degree-2  1  1  2m , ,  dx   dx , ,  dx , , [Supp-16] [Supp-18] [Pu Board Q.B-2m], , Solution:, 3., , Order is 3 and degree is 1  1  1  2m , , Find the order and degree of the differential, 2, , equations xy, , d2y, dy,  dy ,  x   y, 0, 2, dx, dx,  dx , , [Annual-14] [Pu Board Q.B-2m], , Solution: Order -2, Degree-1  1  1  2m , 1

Page 2 : 4. Find the order and degree of the differential, 2, equations if defined  dy   dy  sin 2 y  0,  dx  dx, [Annual-17] [Pu Board Q.B-2m], , Order-1 and degree -2  1  1  2m , 5. Find the order and degree of the equation (if, Solution:, , 4, 3, defined) d y  sin  d y   0, 4, 3, , [Annual-16] [Supp-17] [Pu Board Q.B-2m], , Solution: order=4 and degree is not defined.,  1  1  2m , 6. Find the order and degree of the differential, 2, , 2, equations if defined  d y   cos dy   0,  2, ,  dx , ,  dx , , [Supp-15] [Annual-19] [Pu Board Q.B-2m], , Solution: Order =2 and degree is not defined.  1  1  2m, 7. Find the order and degree of the differential, 3, , equations if defined  d y    dy   sin dy   1  0,  2, 2, , 2, ,  dx , ,  dx , ,  dx , , [Annual-15] [Pu Board Q.B-2m], , Solution:Order-2 and degree doesn’t exist  1  1  2m , 8. Find the order and degree of the differential, 2, equations if defined  ds   3s  d 2s   0, 4, ,  dt , , SOLUTION OF A DIFFERENTIAL EQUATION, , A solution of a differential equation is a function in the, variable involved in the differential equation, which, satisfies the given differential equation., GENERAL SOLUTION OF A DIFFERENTIAL EQUATION, ,  dx , , dx, , 100% SUCCESS FOR II-PU BOARD EXAMINATION, TOPIC-2-GENERAL AND PARTICULAR SOLUTION OF A, DIFFERENTIAL EQUATIONS, ,  dt , , A solution of a differential equation which contains as, many arbitrary constants as the order of the differential, equation is called the general solution or primitive, solution of the differential equation., 2, Example: the general solution of d y  y  0 is, dx 2, y  A cos x  B sin x, PARTICULAR SOLUTION OF A DIFFERENTIAL, EQUATION, , The solution of a differential equation obtained by, giving particular values to the arbitrary constants in the, general solution is called the particular solution., 2, Example: The general solution of d y  y  0 is, 2, , dx, y  A cos x  B sin x but if A  B  1 then, y  cos x  sin x is a particular solution of the given, differential equation, , [Pu Board Q.B-2m], , SOLVED PROBLEMS, , Solution:Order-2 and degree-1  1  1  2m , 9. Find the order and degree of the differential, equations, ,  y, , 2, ,   y    y   y 5  0, 3, , 4, , [Annual-18] [Pu Board Q.B-2m], , Order=3 and degree=2  1  1  2m , 10. Find the order and degree of the differential, ( ), equation, ., Solution:, , [Pu Board Q.B-2m], , Solution: Order-2 and degree is not defined, 11. Find the order and degree of the differential, ( ), equation, ., [Supp-19], , Solution: Order-3 and degree doesn’t exist, 12. Find the number of arbitrary constants in the, general solution of differential equation of fourth, order also find the number of arbitrary constants in, the particular solution of differential equation of, third order. [Pu Board Q.B-2m], Solution: i)4 ii)0, , TWO MARKS QUESTIONS, , 1. Verify the function y  x  2 x  c is a solution of, the differential equation y  2 x  2  0, 2, , [Pu Board Q.B-2m], , y  x 2  2 x  c diff w.r.t. x  1m , , Solution:, , y  2 x  2, , y  2 x  2  0  1m , ,  the given function is a solution of the given, differential equation., , 2. Verify that the function y  1  x 2 is a solution of, xy, the differential equation y ' , 1  x2, [Pu Board Q.B-2m], , Solution: y  1  x 2  1, y , y , y , , 2, , 2x, 2 1  x2, ,  1m , , x 1  x2, 1  x2, , xy, 1  x2, ,  from  1  1m , MCP MATHEMATICS

Page 3 : the given function is a solution of the given, differential equation., 3. Verify that the function y  e x  1 is a solution of, the differential equation y  y '  0, [Pu Board Q.B-2m], , Solution: y  e x  1, Diff w.r.t.x, y  e x  1, Again Diff w.r.t.x, y  e x   2 , y  e x  0, y  y '  0 From-(1), the given function is a solution of the given, differential equation., 4. Verify that the function y  cos x  C is a solution, of the differential equation y ' sin x  0, [Pu Board Q.B-2m], , Solution: y  cos x  C, Diff w.r.t.x, y   sin x  0, y  sin x  0, the given function is a solution of the given, differential equation., 5. Verify that the function y  Ax is a solution of the, differential equation xy '  y,, ,  x  0, , [Pu Board Q.B-2m], , y  Ax, y, A, x, Diff w.r.t.x, xy   y, 0, x2, xy  y  0, the given function is a solution of the given, differential equation., 6. Verify that the function xy  log y  C is a solution, Solution:, , of the differential equation y ' , , y2, 1  xy, , [Pu Board Q.B-2m], , Solution: xy  log y  C, Diff w.r.t.x, , 1, xy  y  y  0, y, Multiplied by y on both sides, , xyy  y 2  y, y  xy  1  y 2  0, CHETHAN M G .SAI PU COLLEGE SHIMOGGA, , DIFFERENTIAL EQUATIONS, , y , , y,  xy  1, 2, , y2, 1  xy , the given function is a solution of the given, differential equation., 7. Verify that the function x  y  tan 1 y is a, solution of the differential equation, y 2 y ' y 2  1  0, y , , [Pu Board Q.B-2m], , Solution: x  y  tan 1 y, Diff w.r.t.x,  1 , 1  y  , y, 2 ,  1 y , , 1  y 1  y 2   y, , 1  y 2  y   yy 2  y , y 2 y ' y 2  1  0, the given function is a solution of the given, differential equation., , 8. Verify that the function y  a 2  x 2 is a solution, dy, of the differential equation x  y, 0, dx, [Pu Board Q.B-2m], , Solution: y  a 2  x 2  1, Diff w.r.t.x, , dy, 2 x, , dx 2 a 2  x 2, dy  x From-(1), , dx, y, dy, y  x, dx, dy, x y, 0, dx, THREE MARKS QUESTIONS, , 1. Verify that the function y  x sin x is a solution of, the differential equation xy '  y  x x 2  y 2, ,  x  0, and, , x  y or x   y , , [Pu Board Q.B-3m], , Solution: y  x sin x  1, Diff w.r.t.x, y  x cos x  sin x, Multiplied by x on both sides, xy  x 2 1  sin 2 x  x sin x, 3

Page 4 : 100% SUCCESS FOR II-PU BOARD EXAMINATION, , y, and from (1) y  x sin x  sin x , x, xy  x 2 1 , , y2, y, x2, , xy  x 2, , x2  y 2, y, x2, , x 2, xy , x, , x y y, 2, , xy  y  x x 2  y 2, the given function is a solution of the given, differential equation., 2. Verify that the function, y  a cos x  b sin x where a, b  R is a solution of the, , d2y, y0, dx 2, , Solution: y  a cos x  b sin x where a, b  R, y   a sin x  b cos x  1m , y  a cos x  b sin x   y  1m , d2y,  y  0  1m , dx 2, the given function is a solution of the given, differential equation., 3. Verify that the function y  cos y  x is a solution, y  y  0 or, , of the differential equation  y sin y  cos y  x  y  y, [Pu Board Q.B-3m], , Solution:, , Given y  cos y  x  1, , Diff w.r.t.x, y   sin y  y  1   2 , LHS   y sin y  cos y  x  y, , From-(1) y  cos y  x, LHS   y sin y  y  y, LHS   y sin y  y y From-(2), LHS  1 y  y, LHS  RHS, , the given function is a solution of the given, differential equation., 4. Verify that the function xy  ae x  be x  x 2 is a, solution of the differential equation, d2y, dy, x 2  2  xy  x 2  2  0, dx, dx, [Pu Board Q.B-3m], , Solution: xy  ae  be, x, , Diff w.r.t.x, 4, , d 2 y dy dy,  ,  xy  x 2  2, 2, dx, dx dx, d2y, dy, x 2  2  xy  x 2  2  0, dx, dx, 5. Verify that the function y  e x  a cos x  b sin x  is, x, , 2, , differential equation, , dy,  y  ae x  be  x  2 x, dx, Diff w.r.t.x, 2, d y dy dy, x 2  ,  ae x  be x  2, dx, dx dx, From (1)  xy  x 2  ae x  be  x, x, , x, ,  x  1, 2, , a solution of the differential equation, d2y, dy,  2  2y  0, 2, dx, dx, [Pu Board Q.B-3m], , Solution: y  e x  a cos x  b sin x   1, Diff w.r.t.x, , dy,  e x  a sin x  b cos x   e x  a cos x  b sin x    2 , dx, dy, from (1),  e x  a sin x  b cos x   y   3, dx, dy,  y  e x  a sin x  b cos x    3, dx, Diff w.r.t.x, 2, d y dy,   e x  a cos x  b sin x   e x  a sin x  b cos x , dx 2 dx, d 2 y dy, dy, ,  e x  a cos x  b sin x  ,  y from-(3), 2, dx, dx, dx, d 2 y dy, dy, from-(1), ,  y , y, 2, dx, dx, dx, d 2 y dy, dy, ,  y , y, 2, dx, dx, dx, d 2 y dy, dy, ,  y, y0, dx 2 dx, dx, d2y, dy, 2,  2y  0, 2, dx, dx, 6. Verify that the function y  x sin 3x is a solution of, 2, , the differential equation d y2  9 y  6 cos 3x  0, dx, [Pu Board Q.B-3m], , Solution: y  x sin 3x  1, Diff w.r.t.x, , dy,  3x cos 3 x  sin 3 x   2 , dx, Diff w.r.t.x, MCP MATHEMATICS

Page 5 : DIFFERENTIAL EQUATIONS, 2, , d y,  3x  3sin 3x   3cos 3x  3cos 3 x, dx 2, d2y,  9 x sin 3x  6 cos 3x   3, dx 2, d2y,  9 y  6 cos 3x From-(3&1), dx 2, LHS  9x sin3x  6cos3x  9x sin3x  6cos3x, LHS  0, LHS  RHS, , LHS , , TOPIC-3-FORMATION OF A DIFFERENTIAL EQUATION, WHOSE GENERAL SOLUTION IS GIVEN, , a) If the general solution contains only one arbitrary, constant i.e., F1  x, y, a   0  1, b) Differentiate equation 1 w.r.t. ‘x’, g x, y, y, a   0, ,  2, , c) The required differential equation is obtained by, eliminating ‘a’ from equation 1 & 2 i.e.,, F x, y, y   0, , 1., a) If the general solution contains two arbitrary, constant i.e., F1 x, y, a, b  0,  1, b) Differentiate equation 1 w.r.t. ‘x’, g x, y, y, a, b  0  2, c) Differentiate equation 2  w.r.t. ‘x’, , g x, y, y, y, a, b  0, ,  3, , d) The required differential equation is obtained by, eliminating ‘a’& ‘b’ from equation 1, 2 & 3 i.e.,, F x, y, y, y  0, Note: the order of a differential equation representing a, form a curves is same as the number of arbitrary, constants in the equation corresponding to the family of, curve, SOLVED PROBLEMS, TWO MARKS QUESTIONS, , 1. From the differential equation of the family of, parabolas having vertex at origin and axis along, positive direction of x  axis .[Pu Board Q.B-2m], Solution: y 2  4ax  y  4a , 2, , x, , 2 yy   4a  1m , 2 yy , , , , y2, x, , OR 2 yx  y  0  1m , CHETHAN M G .SAI PU COLLEGE SHIMOGGA, , THREE MARKS QUESTIONS, , 1. Form the differential equation representing the, family of curves y  mx where m is arbitrary, constant. [Annual-14], Solution: y  mx, y   m  1m , , y , , , y, x, , y   1m , ,  m x , , , , yx  y  0  1m , , 2. Form the differential equation representing the, family of curves y  a sin x  b where a & b are the, arbitrary constant. [PUB.Q.B-3M] [Supp-15], [Annual-15] [Annual-18][Supp-18], , Solution: y  a sin  x  b   1, , y  a cos  x  b   1m , , y  a sin  x  b   1m , y   y  From 1, y   y  0  1m , , 3. Form the differential equation whose general, solution is y  ae3 x  be 2 x by eliminating, arbitrary constant., [PUB.Q.B-3M] [Annual-19] [Supp-19], , Solution: y  ae  be, 3x, , 2 x, , y  3ae3 x  2be2 x  1  1m , y  9ae3 x  4be2 x   2   1m , eq  2   eq 1, , , , , , y  y  6ae3 x  6be2 x  6 ae3 x  be2 x  6 y, y  y  6 y  0  1m , , 4. Form the differential equation representing the, family of curves y  e 2 x a  bx, , Solution: y  e2 x  a  bx   1, , [PUB.Q.B-3M], , diff w.r.t.x, , y  e2 xb  2e2 x  a  bx   1m , , y  be2 x  2 y  from  1 diff w.r.t.x   2   1m , , y  2be 2 x  2 y, y  2  y  2 y   2 y  from  2 , , y  4 y  4 y  1m , , 5. Form the differential equation representing the, family of circles having centre on Y-axis and radius, 3units.[PUB.Q.B-3M] [Annual-17], 5

Page 6 : 100% SUCCESS FOR II-PU BOARD EXAMINATION, , Solution:, ,  x2   y  k   9, 2, , . (0,k), , 7. Form the differential equation representing the, family of circles touching the Y-axis at origin., [PUB.Q.B-3M], , Solution:, , 3, ,   x  a   y2  a2, 2, , ., , (a,0), a, , x 2   y  k   32  1  1m , 2, , 2 x  2  y  k  y  0  1m , ,  x  a, , x   y  k  y  0, , x 2  y 2  2ax, x2  y 2,  2a diff w.r.t.x, x, , x,   2, y, , , , x 2  y    x   9  y , ,  y ,  y , , 2, , 2, , x, , , 2, , 2, ,  9   x2, , 2, , x, 2 x 2  2 xyy  x 2  y 2  0, , 2, , x, , 2, , 9  x , 2, , x,  dy , Or   , 9  x2,  dx , 2, , , ,  1m , , , , 6. Form the differential equation representing the, family of circles touching the X-axis at origin., [PUB.Q.B-3M] [Supp-14] [Annual-15], , Solution:, ,  x2   y  a   a2, 2, , y , , y 2  x2,  1m , 2 xy, , 8. Form the differential equation representing the, x y, family of curves   1 , where a and b are, a b, constant. [PUB.Q.B-3M] [Supp-16], x y,   1 Differentiate w.r.t .x., a b, 1 1,  y   0  1m , a b, , Solution:, , .(0,a), a, , y  , , b,  1m , a, , y  0, x2   y  a   a 2  1m , 2, , x 2  y 2  a 2  2ay  a 2, , x 2  y 2  2ay, x2  y 2,  2a diff w.r.t.x, y, , , , , , y  2 x  2 yy   x 2  y 2 y , y2, 2 yx  2 y 2 y  x 2 y  y 2 y  0, , , , , , 2 yx  y 2 y 2  x2  y 2  0, , , , , , 2 yx  y y  x, , 2, , 2 yx, y  2, x  y2, ,  1m , , 6, , 2, , 0, ,  1m , , x 2  2 xyy  y 2  0, , , , 2, ,  0, , x  2 x  2 yy   x 2  y 2, , 2, ,  x, Put  2  in 1  x      9,  y , 2, , 2, ,  y 2  a 2  1m , , x 2  y 2  a 2  2ax  a 2, ,  y  k  y   x,  y k  , , 2, ,  1m , , 9. Form the differential equation representing the, family of ellipses having foci on x-axis and centre, at origin. [PUB.Q.B-3M], 2, 2, Solution: x 2  y2  1, a, b, 2x 2 y, , y  0, a 2 b2, ,  0  1m , , yy, b2,   2  1m , x, a, , , , x yy   y , , , , x, , 2, , 2, ,   yy  0  1m, , , , x yy   y   yy   0, 2, , xyy  x  y  yy  0  1m , 2, , MCP MATHEMATICS

Page 7 : DIFFERENTIAL EQUATIONS, , METHODS OF SOLVING FIRST ORDER,, FIRST DEGREE DIFFERENTIAL EQUATIONS, ,  tan y , log ,  log c, x ,  1 e , , TOPIC-4-SOLUTION OF THE DIFFERENTIAL, EQUATIONS BY VARIABLE SEPARABLE METHOD, ,  tan y , c, , x ,  1 e , , A first order first degree differential equation is of the, form dy  F  x, y ,  1, dx, dy,  g x   h y ,  2, dx, If h  y   0 separating the variable in equation 2 , can be written as 1 dy  g  x dx, h y , On integrating both sides, we get, , tan y  c 1  e x, ,  h y  dy   g  x dx, 1, , H  y   G  x   c is Required solution of given, differential equation., SOLVED PROBLEMS, , , , , , , , 3. Solve, , , , f   x  dx, f  x, , ,  log f  x  , , , dy,  1  x  y  xy .[PUB.Q.B-3M], dx, , Solution: dy  1  x  y  xy  1  x   y 1  x , dx, dy,  1  x 1  y , dx, , dy,  1  x  dx, 1  y , dy, ,  1  y    1  x  dx  1m, log 1  y   x , , THREE MARKS QUESTIONS, , 1  y   e, , 1. Find the general solution of the differential, equation dy  1  y  [PUB.Q.B-3M], 2, 2, , ye e, c, , 1  x , dy 1  y , , dx 1  x , dx, ,  y  Ce, , 2, , Solution:, , 2, , dy, , 2, , , , dx,  1m , 1  x2, , tan 1 y  tan 1 x  c  1m , 2. Find the general solution of the differential, equation e x  tan y  dx  1  e x  sec2 y  dy  0, , , , [PUB.Q.B-3M], , , , , , , , , , Dividing both sideby tan y  1  e x, 2, , e, sec y, dx , dy  0, x, tan y, 1 e, , , , , , e, , x, , x2, x, 2, , x2, c, 2, ,  1m , , 2, , x, 2, , 1, ,  1  1m , , 2, , sec y, ,  1  e  dx   tan y dy  0, , dy, 1, , dx x x 2  1, , , , 1, ,  dy   x  x, , , , , , , , , ,  log 1  e x  log  tan y   log c, CHETHAN M G .SAI PU COLLEGE SHIMOGGA, , 2, , , , 1, , dx, , , ,  dy   x  x  1 x  1 dx  1, 1, A, B, C,  , , x  x  1 x  1 x x  1 x  1, , 1  A  x  1 x  1  Bx  x  1  Cx  x  1, Put x  1  1  2C  C , , x, ,  sec 2 y , e x, , dx, ,   tan y  dy  0, 1  ex, , , 1, , Solution: e x  tan y  dx  1  e x  sec2 y  dy  0, , x, , x, , x, , x2, c, 2, , 4. Find the particular solution of the differential, equation x  x 2  1 dy  1; y  0 when x  2, dx, dy, Solution: x  x 2  1  1, dx, , dy, dx, ,  1m , 2, 1 y, 1  x2, ,  1 y, , , , 1, ., 2, , 1, 2, Equating the coefficient of, Put x  1  1  2 B  B , , x2  A  B  C  0 , , 1, 1,  B   0  B  1, 2, 2, , 7

Page 8 : 100% SUCCESS FOR II-PU BOARD EXAMINATION, , B, C , A,  dy    x  x  1  x  1  dx  1, 1, 1, 1, 1, 1,  dy   x dx  2   x 1 dx  2   x  1 dx, , 1, 1, y   log x  log x  1  log x  1  log k, 2, 2, 1, y   2log x  log x  1  log x  1  2log k , 2, ,  k 2  x  1 x  1 , 1, y  log ,    2, 2 , x2, , Put x  2, y  0 in  2 , ,  k 3, k 3, 4, 0  log , 1  k2 ,  , 4, 3,  4 , 4, put k 2  in equation-2 we get required solution, 3, 2, , 2, ,  4  x  1 x  1 , 1, y  log , , 2, 3x 2, , , ,  y 3  3x 2  15, , , , y  3 x 2  15, , Solution:, , , , curve at any point is, , , , , , , , , , dy,  1  x2 1  y2, dx, dy,  1  x 2 dx  1m , 1 y2, dy, 2,  1  y 2  1  x dx, , , , tan 1 y  x , , , , x3,  c  1m , 3, , 6. Find the equation of the curve passing through the, point  2,3 given that the slope of the tangent to, the curve at any point is 2x2 ., y, [PUB.Q.B-3M] [Supp-17], , dy 2 x, Solution:, , dx y 2, y 2 dy  2 xdx, ,  y dy   2xdx, 2, , y3, x2,  2  C  1  1m , 3, 2, , Put  2,3 in 1, 27,  4  C  C  9  4  5  1m , 3, , Equation -1 , 8, , x, ., y, , [PUB.Q.B-3M] [Annual-16], , Solution:, , dy x, , dx, y,  ydy  xdx, ,  ydy   xdx  c, y 2 x2,   c  1  1m , 2, 2, Put x  1& y  1 in 1, 1 1,  c c 0, 2 2, , Put c  0 in 1 we get required equation, , dy,  1  x2 1  y2, dx, , , ,  1m , , 7. Find the equation of the curve passing through the, point 1,1 given that the slope of the tangent to the, , 5. Find the general solution of the differential, equation, , , , 1, 3, , y3 2,  x 5, 3, , y 2 x2, , 2, 2, ,  y 2  x 2  0  1m , 8. Find the equation of a curve passing through the, point (0, –2) given that at any point (x, y) on the, curve, the product of the slope of its tangent and y, coordinate of the point is equal to the x coordinate, of the point. [PUB.Q.B-3M], Solution:, , y, , dy,  x  1m , dx, , ydy  xdx, ,  ydy   xdx, y2 x2,   c  1m , 2, 2, 2, 4 0,  cc  2, 2 2, , Put  x, y    0, 2 , , y 2 x2,   2 OR y 2  x 2  4  1m , 2, 2, 9. Find the particular solution of the differential, equation dy  4 xy 2 given that y=1, when x= 0., dx, [PUB.Q.B-3M], , Solution: dy  4 xy 2, dx, , dy,  y 2  4 xdx, dy,  y 2  4 xdx, , 1, 4 x2, , c, y, 2, MCP MATHEMATICS

Page 9 : DIFFERENTIAL EQUATIONS, , Put x  0, y  1  c  1, , 12. In a bank, principal increases continuously at the, rate of 5% per year. An amount of Rs 1000 is, deposited with this bank, how much will it worth, , 1,  2 x 2  1, y, , y, , 2x2  1, 10. Find the particular solution of the differential, equation 1  e2 x  dy  1  y 2  e x dx  0, when y  1 and x  0 [PUB.Q.B-3M], , dy, ex, , dx  0  1m , 2, 1 y, 1  e2 x, dy, , , , e, , x, ,  1  y   , 1   e , 2, , x, , 2, , , , dx  0, , , 4, , , , , 4, , cc, , , Put c in eq-1 tan y  tan e , 2, 1, , , , , , 1, , x, , , 2, ,  1m , ,  1m , , ,  f ( x) ,   f ( x)  dx  log  f ( x)  c , , 11. In a bank, principle P increases continuously at the, rate 5% per year. In how many years Rs1000, double itself. [PUB.Q.B-3M], Solution: Let P be a principal amount of any time t, (in year), dP,  5% P, dt, dP, 5, 1, , , P, P, dt 100, 20, dP 1, , dt, P 20, , , , dP, 1, , dt, P  20, , 1, t  c  1, 20, P  1000 when t  0  log1000  c, log P , , 1, t  log1000, 20, put P  2000 and find t, , Eq 1  log P , , log 2000 , , (in year) dP  5% P  1m , dt, , Initially P  1000 when t  0  log1000  c, , tan 1 y  tan 1 e x  c  Put x  0 & y  1  1, tan 1 1  tan 1 1  c , , Solution: Let P be a principal amount of any time t, , dP 1, , dt, P 20, dP, 1,  P   20 dt, 1, log P , t  c  1, 20, , when y  1 and x  0, ,  , ,  [PUB.Q.B-3M], , dP, 5, 1, , P, P, dt 100, 20, , Solution: 1  e2 x  dy  1  y 2  e x dx  0, , , , , , after 10 years e0.5  1.648, , 1, , 1, t  log1000, 20, ,  2000 , 20 log , t,  1000 , , t  20log  2 , CHETHAN M G .SAI PU COLLEGE SHIMOGGA, , Eq 1  log P , , 1, t  log1000  1m , 20, , put t  10 and find p, log P , , 1, 10   log1000, 20, ,  P , log ,   0.5,  1000 , ,  P  0.5, , e,  1000 , ,  , ,  P  1000 e0.5  1000 1.648,  P  1648  1m , TOPIC-5-HOMOGENEOUS DIFFERENTIAL EQUATIONS, HOMOGENEOUS FUNCTION, , A function F  x, y   h x, y  is said to be, g  x, y , homogeneous function of degree ‘n’ if, F x, y   n F x, y  & n  N for any non-zero, constants, Example:, i) F  x, y   cos y , x,  y ,  y,  y, F x, y   cos   cos   0 cos   0  F  x, y , , x, x,  ,  , x, , Hence it is homogeneous function of degree ‘0’, ii) F x, y   y 2  2 xy, , F x, y   y   2x y   2 y 2  22 xy  2 F  x, y , 2, , Hence it is homogeneous function of degree ‘2’, iii) F  x, y   sin x  sin y, F x, y   sin x  sin y  n F x, y  n  N, Hence it is not a homogeneous function, 9

Page 10 : 100% SUCCESS FOR II-PU BOARD EXAMINATION, , iv) F  x, y   hx, y , g  x, y , , 2. Show that the differential equation x  y , ,  y, x n h ,  x  f y, F  x, y  ,  ,  y, x, x n g , x,  y,  y, n x n h  0 h , x ,  x, F x, y  ,  y,  y, n x n g   g  , x,  ,  x, 0, F x, y    F x, y , This shows that F  x, y  is a homogeneous function, of degree ‘0’, HOMOGENEOUS DIFFERENTIAL EQUATION, , A differential equation of the form, dy, h x, y ,  F  x, y  , dx, g  x, y , , is said to be homogeneous if F  x, y  is a, homogeneous function of degree zero, SOLUTION OF HOMOGENEOUS DIFFERENTIAL, EQUATION, , dy, h  x, y ,  y,  F  x, y   f    1 whereF  x, y  , dx, g  x, y , x, , put, , y  vx , , diff, , w.r.t x, , y, v, x, , dy, x dv, v,   2, dx, dx, dv, x  f v   v, dx, dy, dx, , f x   v x, , dy, , dx,   3, x, Solve equation 3 we get solution of the, differential equation 1, Note: if the homogeneous differential equation is in the, form dx  F  x, y  then, dy, i) we make substitution x  vy , , x, v, y, , ii) we proceed further to find the general solution as, discussed above, SOLVED PROBLEMS, TWO MARKS QUESTIONS, , 10, , is homogeneous. [Pu Board Q.B-2m], Solution:  x  y  dy  x  2 y, dx, , , dy x  2 y, ,  F  x, y   1m , dx  x  y , , F   x,  y  , ,  x  2y ,  x  2 y,  0 , , x   y,  x y , , F   x,  y    0 F  x, y   1m , , Hence given differential equation is homogeneous., THREE MARKS QUESTIONS, , 1. Solve the homogeneous differential equation,  y  dy,  y, x cos    y cos    x, x, dx,  , x, y dy, y, Solution: x cos    y cos    x,  x  dx,  x,  y, y cos    x, dy, x, , dx,  y, x cos  , x, y, dy, dv, put  v  y  vx ,  1m , vx, x, dx, dx, dv vx cos  v   x, vx , dx, x cos  v , , x, ,  f  x  v  , , dy,  x  2y, dx, , x, , dv vx cos  v   x, , v, dx, x cos  v , , dv vx cos  v   x  vx cos  v , , dx, x cos  v , , dv, 1, , dx cos  v , dx, cos  v  dv , x, dx,  cos  v  dv   x  c  1m, sin  v   log x  c, x, ,  y, sin    log x  c  1m , x, 2. Solve the differential equation:,  dy , x 2 .    x 2  2 y 2  xy,  dx , Solution: x 2 .  dy   x 2  2 y 2  xy,  dx , MCP MATHEMATICS

Page 11 : DIFFERENTIAL EQUATIONS, SOLUTION OF LINEAR DIFFERENTIAL EQUATIONS, , dy x 2  2 y 2  xy, , dx, x2, , i) a) dy  Py  Q  1, dx, , 2, , dy,  y  y,  1 2    , dx, x x, , b) Find the integrating factor (I.F) by, , I.F  e , , put y  v  y  vx  dy  v  x dv  1m , x, , dx, , vx, , c) write the solution of the given differential, equation is yI .F    QI .F dx  C, , dx, , dv, 2,  1 2 v  v, dx, x, , ii) a) dx  Px  Q  1, dy, , dv, 2,  1 2 v, dx, , dv, ,  1 2 v  , 2, , b) I.F  e, c) xI .F    QI .F dy  O, Pdy, , dx, x, , 1, dv, dx, , 2, , 2  1 , x, 2, ,   v,  2, 1, v, 1, 1, log 2,  c  log x, 1, 2  1 , v, 2, , 2,  2, 1, y, , 1, 1, x, 2, log x , log, c, 1, y, 2  1 , , 2, , 2 x,  2, , log x , , 1, 2 2, , log, , SOLVED PROBLEMS, FIVE MARKS QUESTIONS, , 1. Solve the differential equation, x  log x  dy  y  2 log x  [PUB.Q.B] [Supp-14], dx, x, dy, 2, Solution: x  log x   y  log x , dx, x, [Divided by  x  log x  on both sides],  dy, ,  dx  Py  Q   1m , , , 1, 2, P, & Q  2  1m , x, , log, x, x, , , , dy, y, 2, ,  2, dx  x  log x  x, , x  2y, c, x  2y, , TOPIC-6-SOLUTION OF THE FIRST ORDER FIRST, DEGREE LINEAR DIFFERENTIAL EQUATIONS, , I .F  e, , e, , , ,  1m , , 1  2 ,  2 , y  log x   log x       dx  c, x x ,  x , , 2, 2, , , y  log x   log x ,    2 dx  c, x,  x ,  2  2, y  log x   log x ,  c,  x  x,  2  2, y  log x   log x ,  c,  x  x, , ii) Another form of first order first degree liner, , dy,  Px  Q where P &, dx, , Q are constant or function of y only, Example: i) dx  x  cos y, dy, , y, , , , 2, , y  log x     2  log x  dx  c  1m , I, x, ,  II, , , dx, ii) dy   1  y  e x, dx  x , , dy, , e, , 1, x dx,  log x , , y  I .F    Q  I .F  dx  c, , Example: i) dy  y  sin x, , ii) dx   2 x  y 2 e  y, , P dx, , 1, dx,  xlog x , ,  eloglog x  log x, , dy, i) A differential equation of the form,  py  Q, dx, where ‘P’&’Q’ are constant or function of ‘x’ only, is known as a first order first degree liner, differential equation, , differential equation is, , Pdx, , [Divided by, , y, , 2., ,  log x  on both sides], , 2, 2, c,  1m , , , x x  log x logx, , Find the general solution of the differential, dy, equation x  2 y  x 2  x  0 , dx, [PUB.Q.B] [Supp-16] [Supp-18] [KCET-2015-1M], , CHETHAN M G .SAI PU COLLEGE SHIMOGGA, , 11

Page 12 : 100% SUCCESS FOR II-PU BOARD EXAMINATION, , [KCET-2017-1M], , Solution: x dy  2 y  x 2 [Divided by x on both sides], dx, dy 2,  y  x  dy  Py  Q   1m , dx x,  dx, , , P, , 2, & Q  x  1m , x, , I .F  e , , Pdx, , e, , , , 2, dx, x, ,  elog f  x   f  x  , , , 2, , y  I .F    Q  I .F  dx  c, , yx 2 , , 4, , [Divided by x 2 on both sides], , x2,  cx 2  1m , 4, , [PUB.Q.B] [Annual-15] [Annual-19], , dy, ,  y  sec x  tan x, 0  x , dx, 2, P  sec x & Q  tan x  1m , , Solution:, , Pdx, sec xdx, I .F  e ,  e,  elog secx  tan x ,  secx  tan x  1m , ,   , , , ,   tan x  e, , tan x, , y  sec x  tan x    tan x  sec x  tan x dx  c, , y  sec x  tan x     tan x sec x  tan 2 x  dx  c,    tan x sec x  sec2 x  1 dx  c  1m , , y  sec x  tan x   sec x  tan x  x  c  1m , 4. Solve the differential equation, dy, , , cos 2 x,  y  tan x  0  x  , dx, 2, , [PUB.Q.B] [Annual-17], , cos 2 x, , , , dy,  y  tan x   cos2 x , dx, , , , , , , , dy,  sec2 x y  tan x sec 2 x  1m , dx, , , , P  sec2 x & Q  tan x sec2 x, , I .F  e , , sec 2 x dx, ,  e tan x  1m, , y  I .F    Q  I .F  dx  c, , , , tan x, ,  etan x  c  1m , , 5. Find the particular solution of the differential, dy, equation,  y  cot x  2 x  x 2  cot x given that, dx, , , 2, , [PUB.Q.B], , Solution: dy  y  cot x  2 x  x 2  cot x, dx, ,  P  cot x & Q  2 x  x 2 cot x  1m , Pdx, ,  e, , cot xdx, ,  elog sin x   sin x  1m , , y  I .F    Q  I .F  dx  c  1m, , , , , , y  sin x    2 x  x2 cot x  sin x dx  c, , , , I, , , ,   , , , , y  sin x    2 x sin x  x 2 cos x dx  c, II, , , , , , y  sin x   sin x x2   x2 cos x dx   x 2 cos x dx  c, y sin x  x 2 sin x  c  1  1m , Put x , , , 2, , & y 0c , , 2, 4, , ,  1m , 4sin x, 6. Find the particular solution of the differential, dy, equation,  3 y  cot x  sin 2 x ,, dx, , when y  2 and x  [PUB.Q.B], 2, eq 1  y  x 2 , , y  I .F    Q  I .F  dx  c  1m , , 12, , , , y etan x  t et  et  c, , I .F  e, , 3. Solve the differential equation, dy, ,  y  sec x  tan x, 0  x , dx, 2, , Solution:, , , , , , y etan x   et t  dt  c  1m , , 3, , x, c, 4, , y, , , , , , y  0 when x , , yx   xx dx  c   x dx  c  1m , 2, , , , y e, ,  e2log x  elog x  x2  1m , 2, , , , y e tan x   e tan x tan x  sec 2 x  dx  c  1m , Put tan x  t  sec2 xdx  dt, , 2, , Solution: dy  3 y  cot x  sin 2 x  dy  Py  Q ,  dx, , dx, , , P  3cot x, Q  sin 2 x, , I .F  e , , P dx, , 3 cot x dx, e ,  e 3log sin x  , , y  I .F    Q  I .F  dx  c, , 1, sin 3 x, , 1 ,  1 , , y  3    sin 2 x  3  dx  c, sin x ,  sin x , , 1 ,  1 , , y  3   2  sin x  cos x  3  dx  c, sin, x, sin, x, , , ,  1 ,  sin x  cos x , y  3   2  , 3,  dx  c,  sin x ,  sin x , , MCP MATHEMATICS

Page 13 : DIFFERENTIAL EQUATIONS, ,  1 , y  3   2  sin 2 x cos x  dx  c,  sin x ,  1  2  sin x , y 3  , 2  1,  sin x , 2,  1 , y 3   , c, sin x,  sin x , , 2 1, , c  , , , , , , , , y 1  x 2  tan 1 x , , 2, c, sin x, , 3, , 7. Find the particular solution of the differential, , dx, , I .F  e , ,  1 x2 dx, , e, , , , log 1 x 2, , , , y 1 x, , , , , ,  1  x2, ,   1m, , x3,  c  1, 3, ,  1m ,  1m , , [PUB.Q.B], , Solution: 1  x 2  dy  2 xy , dx, , 1, 1  x2, , dy, 2 xy, 1, , , dx 1  x 2, 1  x2, , , , I .F  e, , , ,  , , , , 2x, , P dx, , e, ,  1 x2, , dx, , , , e, , , , , , 2, , , , log 1 x2, , , , , , , , , , , , , , , , , , log xdx  c, , , , equation x dy  2 y  x 2 log x, dx, , [PUB.Q.B] [Annual-18], , Solution:, , x, , dy,  2 y  x 2 log x, dx, , dy 2,  y  x log x, dx x, 2,  P  & Q  x log x, x, , I .F  e , , 2, ,  1m , , , 1, y 1  x2   , 1  x2,  1  x2 2, , 1, 2, y 1 x  , dx  c, 1  x2, , , ,  1m , , 10. Find the general solution of the differential, , dy, 1,  2 xy , , y  0 when x  1, dx, 1  x2, , 2x, 1, P, ,Q , 1  x2, 1  x2, , 3, ,  , , Equation (1)  y 1  x 2   x  x  4, 3 3, 8. Solve the differential equation, , , , 2, ,  1m , ,  , , 3, , 1  x , , 2, ,  x4  x4, y x 2  log x     c,  4  16, 4, x, y x 2   4log x  1  c, 16, x2, y   4 log x  1  cx 2  1m , 16, , put y  0 & x  1in  1, , 2, ,  e 2log x   elog  x   x 2, ,  , ,    1  x dx  c, , 1, 4, 0  1  c  c  , 3, 3, , 2, ,  x dx, ,  , , 2, , y 1  x 2   x , , e, ,  1m , ,  x4  1  x4 , y x 2  log x       dx  c, x 4 ,  4, 4, x  1, y x 2  log x     x3dx  c  1m ,  4 4, , y  I .F    Q  I .F  dx  c  1m , 2, , x, ,  x, ,  1m , , 2x, , e, , Pdx, , y x, , [PUB.Q.B], , Pdx, ,  1m , , y  I .F    Q  I .F  dx  c, , dx 1  x, , I .F  e , , 4, , dy 2, Solution:,  y  x log x, dx x, 2,  P  & Q  x log x, x, , equation dy  2 xy2  1 when y  0 and x  1, Solution: dy  2 xy2  1, dx 1  x, , , , , ,  c  c    1m , 4, 4, , 9. Find the general solution of the differential, equation dy  2 y  x log x, , , , y  2sin x  c sin x  1  Put y  2 & x  , 2, , , , 2  2sin 2  c sin 3  2  c  c  4, 2, 2, Eq  1  y  2sin 2 x  4sin 3 x, 2, , , , 1, , 0 1  1  tan 1  c  0 , 2, , Pdx, , e, , 2, ,  x dx, ,  1m , ,  e 2log x   e log  x   x 2, 2, ,  1m , , y  I .F    Q  I .F  dx  c  1m , ,  , , y x 2   x3 log xdx  c, ,  1  x 2  1m , , ,  dx  c, , , , , y 1  x2  tan 1 x  c  1  1m , , put y  0 & x  1, CHETHAN M G .SAI PU COLLEGE SHIMOGGA, ,  x4  1  x4 , y x 2  log x       dx  c, x 4 ,  4, 4, x  1, y  x 2   log x     x3dx  c  1m ,  4 4, ,  , ,  x4  x4, y x 2  log x     c,  4  16, 4, x, y  x 2    4log x  1  c, 16, ,  , , y, , x2,  4log x  1  cx 2  1m , 16 , , 13

Page 14 : 100% SUCCESS FOR II-PU BOARD EXAMINATION, , 11. Find the equation of a curve passing through the, point (0, 1). If the slope of the tangent to the curve, at any point (x, y) is equal to the sum of the x, coordinate (abscissa) and the product of the x, coordinate and y coordinate (ordinate) of that point., [PUB.Q.B], , Solution:, , dy, dy,  x  xy ,  xy  x, dx, dx,  P   x & Q  x  1m , ,  xdx, I .F  e , e, ,  x2, 2, ,  1m , , y  I .F    Q  I .F  dx  c  1m , , , ye, , , ,  x2, 2, , ,    xe, , , ,  x2, 2, , , ye, , , , 2, , , t, t,     e dx  c  e  c, , , ,  x2, 2, , ,   e, , , , y  1  c e, ,  x2, 2, , 13. Solve the differential equation, y dx  x  2 y 2 dy  0 ., [PUB.Q.B] [Supp-17], , Solution: y dx  x  2 y 2 dy  0, dx x,   2y, dy y, , 1, ,   dy, 1, Pdy, 1, I .F  e   e y  e  log y  elog y  y 1 , y, , 1, 1, x     2 y   dy  c  2 y  c  1m ,  y,  y, 2, x  2 y  cy  1m , , 14. Find the general solution of the differential, equation  x  3 y 2  dy  y  y  0 , dx, ,  c  1m , , Solution:  x  3 y 2  dy  y  y  0 , dx, , dx,  x  3 y   y dy, , Put  x, y    0,1, , , , divided by y onboth sides , , x, dx,  3y , y, dy, dx x,   3 y  dx  Px  Q , dy y,  dy, , , , ,  1m , , 12. Find the equation of a curve passing through the, point  0, 2  given that the sum of the coordinates of, the point on the curve exceeds the magnitude of the, slope of the tangent to the curve at any point by 5., [PUB.Q.B], , dx, dy,  dy, ,   y  x  5   Py  Q , dx, dx, , , , P  1, Q  x  5  1m , P dx, ,  e   e  x  1m ,  dx, , y  I .F    Q  I .F  dx  c  1m , , ye x    x  5 e x dx  c, , , ,  , , , , ye x   x  5 e x   e x 1 dx  c, , ye x    x  5 e x  e x  c, ,  1m , , Put y  2 & x  0, 2    0  5 e0  e0  c  5  1  c  4  c  2  4  c  c  2, , 14, , 1,  P   & Q  3y, y, ,  1m , ,  1m , , 1, , Solution: dy  5  x  y, , I .F  e , , , , 1,  P   y & Q  2 y   1m , , , , [PUB.Q.B] [Supp-15], , 1  1  c e0  c  2, y  1  2e, , y  4  x  2e x  1m , , 2, , x2, 2, , x2, 2, , y    x  5  1  2e x, , x  I .F    Q  I .F  dy  c  1m , , dx  c, ,  x2, put,  t  xdx  dt, 2, ,  x, ye 2, , , , ye    x  5 e x  e x  2, x, ,   dy, Pdy, 1, I .F  e  e y , y, ,  1m , , x  I .F    Q  I .F  dy  c, 1, 1, x     3 y   dy  c  1m ,  y,  y, , 1, x     3 dy  c  3 y  c,  y, 1, x    3 y  c  multiply by y on both side,  y, , x  3 y 2  cy  1m , , 15. Solve the differential equation, ydx  x  ye y dy  0 [PUB.Q.B] [Annual-16], , , , , , MCP MATHEMATICS, ,  1m 