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.