Differential eq n. 2. Given the solution of. through the point (1, π 4 ) Sol: xsin 2 ( y ) dx = ydx xdy .. (1) By partial fraction + 1

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Claire Harvey
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$required general solution. By partial fraction 2.$

1 Differential eq n. Solve ( 3 3y 2 )d + (3 2 y y 3 ) 0. Sol: ( 3 3y 2 )d + (3 2 y y 3 ) 0. (3 3y 2 ).. () d (3 2 y y 3 ) this is homogenous D. E let y v v + d d Eq n () v + d 3 3(v) (v) (v) 3 v + 3 ( 3v 2 ) d 3 (3v v 3 ) v + d ( 3v2 3v v 3) d +3v2 3v v 3 v +3v2 3v 2 +v 4 d 3v v 3 d ( v4 3v v 3) 3v v3 v 4 d 3v v 3 (v )(v+)(v d 2 +) [ + 2(v+) 2(v ) 2v (v 2 +) ] d 2 log v log v log v2 + log + logc log v+ v log(c) v 2 + v+ v v 2 + c v2 v 2 + v2 (v 2 +) 2 (c)2 c (y2 2 ) c 2 2 (y ) 2 4 (y 2 2 ) (y ) 2 Which is required general solution. By partial fraction 2. Given the solution of sin 2 ( y ) d yd Which passes through the point (, π 4 ) Sol: sin 2 ( y ) d yd yd sin 2 ( y )d [y sin 2 ( y )] d [ y sin2 ( y ) ] d d [y sin2 ( y )].. let y v v + d d v + d v sin2 (v) d sin2 (v) sin 2 (v) d cosec 2 v d cotv log + c cot ( y ) log + c this is passing Through the point (, π 4 ) cot ( π ) log + c c c cot ( y ) log

2 3. solve the differential equation y+3. d 2 2y+5 Sol: d [ a a b b ] y+3 2 2y+5 this is non homogeneous D. E of case(2) y+3.. d 2( y)+5 let ( y) v d d d d Now eq n () becomes d v+3 2v+5 4. solve the diffrential equation (2 + y + )d + (4 + 2y ) 0. Sol: (2+y+) d 4+2y [ a a b b ] this is non homogeneous D. E of case(2) (2+y+) d 4+2y..() let (2 + y) v 2 + d d d d 2 Now eq n () becomes 2 (v+) d 2v v+3 2v+5 d d v 2v + 2 2v+5 v 3 2v+5 v+2 2v+5 d d d 2v+5 v+2 d 2v+4+ v+2 d 2(v+2)+ v+2 d ( 2(v+2) v+2 + v+2 ) v +4v 2 d 2v 3v 3 d 2v 2v 3d v 2(v )+ 3d v ( 2(v ) + ) 3d v v d (2 + v+2 ) 2v + log(v + 2) + c 2( y) + log( y + 2) + c 2y + log( y + 2) c 3d (2 + v ) 3 2v + log(v ) + c 3 2(2 + y) + log(2 + y ) + c y + log(2 + y ) + c + 2y + log(2 + y ) c 2

3 5. solve d 2+y+3 2y++. Sol: Given eq n 2+y+3 d 2y++ [a/a b/b ] this is non homogeneous D. E of case(3) put X + h and y Y + k 2+y+3 d 2y++ 2(X+h)+(Y+k)+3 2(Y+k)+(X+h)+ 2X+Y+(2h+k+3) X+2Y+(h+2k+) ( ) now choose h and k such that 2h + k () and h + 2k + 0 (2) solving ()& (2) (h, k) [ 6 4, ] [ 5 3, 3 ] Hence ( )becomes 2X+Y X+2Y is a homogeneous equation. put y VX XdV V + [ 2(+V) + 3 2( V) ] dv 2 X 2 (+V) dv ( V) dv 2 X log + V 3 log V log X + log C log + V + 3 log V 4 log X + log C log( + V)( V) 3 log(c/x 4 ) ( + V)( V) 3 C X 4 {V Y X } ( + Y X ) ( Y X )3 C X 4 (X+Y)(X Y)3 X 4 C X 4 (X + Y)(X Y) 3 C (h, k) [ 5 3, 3 ] X + 5 3, Y y 3 ( y 3 ) ( y + 3 )3 C ( + y ) ( y + 2)3 C (3 + 3y + 4)( + y + 2) 3 3C This is the required solution. 2X+VX X+2VX X(2+V) X(+2V) (2+V) (+2V) (2+V) (+2V) V 2+V V 2V2 (+2V) 2 2V2 (+2V) 2( V2 ) (+2V) 3

4 6. solve d 3y y 3. Sol: Given eq n 3y 7+7 d 3 7y 3 [a/a b/b ] this is non homogeneous D. E of case(3) put X + h and y Y + k 3y 7+7 d 3 7y 3 3(Y+k) 7(X+h)+7 3(X+h) 7(Y+k) 3 7X+3Y+( 7h+3k+7) 3X 7Y+(3h 7k 3) ( ) now choose h and k such that 7h + 3k () and 3h 7k 3 0 (2) 7+7V2 3 7V 7(V2 ) 3 7V 3 7V V 2 dv 7 X [ 2(+V) + 3 2( V) ] dv 7 X 2 (+V) dv ( V) dv 2 X log + V 3 log V log X + log C log + V + 3 log V 4 log X + log C log( + V)( V) 3 log(c/x 4 ) ( + V)( V) 3 C X 4 {V Y X } ( + Y X ) ( Y X )3 C X 4 solving ()& (2) (h, k) [ , ] [,0] Hence ( )becomes 7X+3Y 3X 7Y is a homogeneous equation. (X+Y)(X Y)3 X 4 C X 4 (X + Y)(X Y) 3 C (h, k) [ 5 3, 3 ] X + 5 3, Y y 3 ( y 3 ) ( y + 3 )3 C (3 + 3y + 4)( + y + 2) 3 3C This is the required solution. put y VX XdV V + 7X+3VX 3X 7VX X( 7+3V) X(3 7V) ( 7+3V) (3 7V) ( 7+3V) (3 7V) V 7+3V 3V+7V2 (3 7V) 4

5 7. solve the diffrential equation +2y+3 d 2+3y+4 Sol: Given eq n +2y+3 [a/a b/b ] d 2+3y+4 this is non homogeneous D. E of case(3) put X + h and y Y + k +2y+3 d 2+3y+4 (X+h)+2(Y+k)+3 2(X+h)+(Y+k)+4 X+2Y+(h+2k+3) ( ) 2X+3Y+(2h+3k+4) now choose h and k such that h + 2k () and 2h + 3k (2) solving ()& (2) V 3V 2 dv X [ 2 + 3V 2 3V 3V 2] dv X 2 dv 6V dv 3 [ 3 ]2 V 2 2 3V 2 X ( 3 +V log 3 ) V log 2 3V2 3 log CX log + 3V log ( 3) 3V 2 3V2 log CX Y X log ) + 3( ( 3) 3( Y X ) 2 log 3 (Y X )2 log CX log X+ 3Y log 2 ( 3) X 3Y 2 X 3Y 2 log CX X 2 X + andy y + 2 ( 3) log + + 3(y 2 ) + 3(y 2) 2 log (+)2 3(y 2) 2 (+) 2 log CX (h, k) [ 8 9, Hence ( )becomes X+2Y 2X+3Y ] [, 2] is a homogeneous equation. put y VX XdV V + X+2VX 2X+3VX X(+2V) X(2+3V) (+2V) (2+3V) (+2V) (2+3V) V +2V 2V 3V2 (2+3V) 3V2 2+3V 5

6 6

Lesson 3: Linear differential equations of the first order Solve each of the following differential equations by two methods.

Lesson 3: Linear differential equations of the first der Solve each of the following differential equations by two methods. Exercise 3.1. Solution. Method 1. It is clear that y + y = 3 e dx = e x is an