Question

Verify that the given function (implicit or explicit) is a solution of the corresponding differential equation.
(i) $y=e^x(a\cos x+bsinx)$ : $\frac{d^{2}y}{d{x}^2}-2\frac{dy}{dx}+2y=0$
(ii) $y=x\sin 3x$ : $\frac{d^{2}y}{d{x}^2}+9y-6\cos 3x=0$
(iii) $x^2=2y^2\log y$ : $(x^2+y^2)\frac{dy}{dx}-xy=0$

Answer 1

(i) $y=e^x(a\cos x+b\sin x)$

Differenttiating Both sides w.r.t.x

$\frac{dy}{dx}=\frac{d}{dx}\left[e^x\right(a\cos x+b\sin x\left)\right]$

$y^{\prime }=\frac{d\left(e^x\right)}{dx}.[a\cos x+b\sin x]+e^x\frac{d}{dx}[a\cos x+b\sin x]$

$y^{\prime }=e^x[a\cos x+b\sin x]=e^x[-a\sin x+b\cos x]$

$y^{\prime }=y+e^x[-a\sin x+b\cos x]$ ...(1)

Again Differentiating both sides w.r.t.x

$y^{\prime \prime }=y^{\prime }+\frac{d\left(e^x\right)}{dx}[-a\sin x+b\cos x]+e^x\frac{d}{dx}[-a\sin x+b\cos x]$

$y^{\prime \prime }=y^{\prime }+e^x[-a\sin x+b\cos x]=e^x[-a\cos x+b(-\sin x\left)\right]$

$y^{\prime \prime }=y^{\prime }+(y^{\prime }-y)+e^x[-a\cos x-b\sin x]$ (From (1)]

$y^{\prime \prime }=2y^{\prime }-y-e^x[a\cos x+b\sin x]$

$y^{\prime \prime }=2y^{\prime }-y-y$ (Using $y=e^x(a\cos x+b\sin x)]$

$y^{\prime \prime }=2y^{\prime }-2y$

$y^{\prime \prime }-2y^{\prime }+2y=0$

Which is the required differential equation

(ii) $y=x\sin 3x;\frac{d^{2}y}{d{x}^2}+9y-6\cos 3x=0$

$\frac{dy}{dx}=\sin 3x+3x\cos 3x$
$\frac{d^{2}y}{d{x}^2}=3\cos 3x+3\cos 3x-9x\sin 3x$

or $\frac{d^{2}y}{dx}=6\cos 3x-9x\sin 3x$

where $x\sin 3x=y$

$\frac{d^{2}y}{d{x}^2}=9y-6\cos 3x=0$

This is the proof that $y=x\sin 3x$ is the solution of the given differential equation.

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

Differentiating both the sides we get;

$2x=4y\log y\frac{dy}{dx}+\frac{2y^2}{y}\frac{dy}{dx}$

$2x=\frac{dy}{dx}(4y\log y+2y)$

$\therefore x=\frac{dy}{dx}(2y\log y+y)$

Multiplying both the sides by y, we get,

$xy=\frac{dy}{dx}$ $(2y^2\log y+y^2)$

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

Hence proved.