Question

For each of the exercises given below, verify that the given function (implicit or explicit) is a solution of the corresponding differential equation. (i) (ii) (iii) (iv)

Answer 1

(i)

Given function is,

y=a e x +b e −x + x 2

Differentiate both sides of equation.

dy dx =a d dx ( e x )+b d dx ( e −x )+ d dx ( x 2 ) dy dx =a e x −b e −x +2x d 2 y d x 2 =a e x +b e −x +2

The differential equation is,

x d 2 y d x 2 +2 dy dx −xy+ x 2 −2=0(1)

Substitute the value of dy dx and d 2 y d x 2 to the Left Hand Side of the above equation.

L.H.S.=x d 2 y d x 2 +2 dy dx −xy+ x 2 −2 =x( a e x +b e −x +2 )+2( a e x −b e −x +2x )−x( a e x +b e −x + x 2 )+ x 2 −2 =( ax e x +bx e −x +2x )+( 2a e x −2b e −x +4x )−( ax e x +bx e −x + x 3 )+ x 2 −2 =2a e x −2b e −x − x 3 + x 2 +6x−2

Further simplify.

L.H.S.=2a e x −2b e −x − x 3 + x 2 +6x−2 ≠0

Left hand side is not equal to right hand side.

Hence, given function is not solution of the corresponding differential equation.

(ii)

Given function is,

y= e x ( acosx+bsinx )

Differentiate both sides of equation.

dy dx =a d dx ( e x cosx )+b d dx ( e x sinx ) dy dx =a( e x cosx− e x sinx )+b( e x sinx+ e x cosx ) dy dx =( a+b ) e x cosx+( b−a ) e x sinx d 2 y d x 2 =( a+b ) d dx ( e x cosx )+( b−a ) d dx ( e x sinx )

Further simplify,

d 2 y d x 2 =( a+b )[ e x cosx− e x sinx ]+( b−a )[ e x sinx+ e x cosx ] = e x { acosx−asinx+bcosx−bsinx+bsinx+bcosx−asinx−acosx } d 2 y d x 2 =[ 2 e x ( bcosx−asinx ) ]

The differential equation is,

d 2 y d x 2 −2 dy dx +2y=0(1)

Substitute the value of dy dx and d 2 y d x 2 to the Left Hand Side of the above equation.

L.H.S.=2 e x ( bcosx−asinx )−2 e x [ ( a+b )cosx+( b−a )sinx ]+2 e x ( acosx+bsinx ) = e x [ ( 2bcosx−2asinx )−( 2acosx+2bcosx )−( 2bsinx−2asinx ) +( 2acosx+2bsinx ) ] = e x [ ( 2b−2a−2b+2a )cosx ]+ e x [ ( −2a−2b+2a+2b )sinx ] =0

Left hand side is equal to right hand side.

Hence, given function is solution of the corresponding differential equation.

(iii)

Given function is,

y=xsin3x

Differentiate both sides of equation.

dy dx = d dx ( xsin3x ) =sin3x+xcos3x×3 =sin3x+3xcos3x d 2 y d x 2 = d dx ( sin3x )+ d dx ( 3xcos3x )

Further simplify.

d 2 y d x 2 =3cos3x+3{ cos3x+x( −sin3x )×3 } =6cos3x−9xsin3x

The differential equation is,

d 2 y d x 2 +9y−6cos3x=0(1)

Substitute the value of d 2 y d x 2 to the Left Hand Side of the above equation.

L.H.S.= d 2 y d x 2 +9y−6cos3x =( 6cos3x−9xsin3x )+9xsin3x−6cos3x =0

Left hand side is equal to right hand side.

Hence, given function is solution of the corresponding differential equation.

(iv)

Given function is,

x 2 =2 y 2 logy

Differentiate both sides of equation.

d dx ( x 2 )=2 d dx ( y 2 logy ) 2x=2{ 2ylogy dy dx + y 2 × 1 y × dy dx } x= dy dx { 2ylogy+y } dy dx = x y( 1+2logy )

The differential equation is,

( x 2 + y 2 ) dy dx −xy=0(1)

Substitute the value of dy dx to the Left Hand Side of the above equation.

L.H.S.=( 2 y 2 logy+ y 2 )× x y( 1+2logy ) −xy = y 2 ( 1+2logy ) x y( 1+2logy ) −xy =xy−xy =0

Left hand side is equal to right hand side.

Hence, given function is solution of the corresponding differential equation.