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) $y=ae^x+be^{-x}+x^2: x\frac{d^2y}{dx^2}+2\frac{dy}{dx}-xy+x^2-2=0$

ii) $y=e^x(a \cos x+ b \sin x):\frac{d^2y}{dx^2}-2\frac{dy}{dx}+2y=0$

iii) $y= x \sin 3x:\frac{d^2y}{dx^2}+9y-6\cos3x=0$

iv) $x^2=2y^2\log y:(x^2+y^2)\frac{dy}{dx}-xy=0$

Answer 1

(i) $y=ae^x+be^{-x}+x^2$

Differentiating both sides with respect to x, we get:

$\frac{dy}{dx}=a\frac{d}{dx}(e^x)+b\frac{d}{dx}(e^{-x})+\frac{d}{dx}(x^2)$

$\Rightarrow \frac{dy}{dx}=ae^x-be^{-x}+2x$

Again, differentiating both sides with respect to x we get:

$\frac{d^2y}{dx^2}=ae^x+be^{-x}+2$

Now ,on substituting the values of $\frac{dy}{dx}$ and$\frac{d^2y}{dx^2}$ in the differential equation, we get:

L.H.S.

x$\frac{d^2y}{dx^2}$+2$\frac{dy}{dx}$-xy+x2-2

=x(α$e^x$+b$e^{-x}$+2)+2(α$e^x$-b$e^{-x}$+2x)-x(α$e^x$+b$e^{-x}$+x2)+x2-2

=(αx$e^x$+bx$e^{-x}$+2x)+(2α$e^x$-2b$e^{-x}$+4x)-(αx$e^x$+bx$e^{-x}$+x3)+x2-2

=2α$e^x$-2b$e^{-x}$+x2+6x-2

≠0

⇒L.H.S.≠R.H.S.

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

(ii) y=$y=e^x(a \cos x+ b \sin x)=ae^x\cos x+be^x\sin x$

Differentiating both sides with respect to x, we get:

$\frac{dy}{dx}$=α.$\frac{d}{dx}$($e^x$ cosx)+b.$\frac{d}{dx}$($e^x$ sinx)

⇒$\frac{dy}{dx}$=α($e^x$ cosx-ex sinx)+b.($e^x$ sinx+$e^x$ cosx)

⇒$\frac{dy}{dx}$=(α+b)$e^x$ cosx+(b-α)$e^x$ sinx

Again, differentiating both sides with respect to x, we get:

$\frac{d^2y}{dx^2}$=(α+b).$\frac{d}{dx}$($e^x$ cosx)+(b-α)$\frac{d}{dx}$($e^x$ sinx)

$\Rightarrow \frac{d^2y}{dx^2}$=(α+b).[$e^x$ cosx-ex sinx]+(b-α)[$e^x$ sinx+ex cosx]

$\Rightarrow \frac{d^2y}{dx^2}=e^x$[(α+b)(cosx-sinx)+(b-α)(sinx+cosx)]

$\Rightarrow$ $\frac{d^2y}{dx^2}$=$e^x$[αcosx-αsinx+bcosx-bsinx+bsinx+bcosx-αsinx-αcosx]

$\Rightarrow \frac{d^2y}{dx^2}$=[$2e^x$(bcosx-αsinx)]

Now ,on substituting the values of d2y/dx2 and $\frac{dy}{dx}$ in the L.H.S. of the given differential equation, we get:

$\frac{d^2y}{dx^2}+2\frac{dy}{dx}$+2y

=$2e^x$(bcosx-αsinx)-2ex[(α+b)cosx+(b-α)sinx]+2ex(αcosx+bsinx)

=$e^x$[(2bcosx-2αsinx)-(2αcosx+2bsinx)-(2bsinx-2αsinx)+(2αcosx+2bsinx)]

=$e^x$[(2b-2α-2b-2α)cosx]+ex[(-2α-2b+2a+2b)sinx]

=0

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

(iii) y=xsin3x

Differentiating both sides with respect to x, we get:

$\frac{dy}{dx}=\frac{d}{dx}$(xsin3x)=sin3x+x.cos3x.3

$\Rightarrow \frac{dy}{dx}$=sin3x+3xcos3x

Again, differentiating both sides with respect to x,we get:

$\frac{d^2y}{dx^2}=\frac{d}{dx}$(sin3x)+3$\frac{d}{dx}$(xcos3x)

$\Rightarrow \frac{d^2y}{dx^2}$=3cos3x+3[cos3x+x(-sin3x).3]

$\Rightarrow \frac{d^2y}{dx^2}$=6cos3x-9xsin3x

Substituting the value of $\frac{d^2y}{dx^2}$ in the L.H.S. of the given differential equation, we get:

$\frac{d^2y}{dx^2}$2+9y-6cos3x

=(6.cos3x-9xsin3x)+9xsin3x-6cos3x

=0

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

(iv)x2=2y2logy

Differentiating both sides with respect to x,we get:

2x=2.$\frac{d}{Dx}$=[y2log y]

$\Rightarrow$ x=[2y.logy.$\frac{dy}{dx}+y^2.\frac{1}{y}$.$\frac{dy}{dx}$]

$\Rightarrow x=\frac{dy}{dx}$(2ylogy+y)

$\Rightarrow \frac{dy}{dx}=\frac{x}{y(1+2logy)}$

Substituting the value of $\frac{dy}{dx}$ in the L.H.S. of the given differential equation,we get:

$(x^2+y^2)\frac{dy}{dx}-xy$

=$(2y^2\log y+y^2).\frac{x}{y(1+2\log y)}-xy$

=$y^2(1+2\log y).\frac{x}{y(1+2 log y)}-xy$

=xy-xy

=0

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