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

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

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

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

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

(i) $y = a{e^x} + b{e^{ - x}} + {x^2}$
Differentiating both sides with respect to ${\rm{x}}$,

we get:
$\frac{{dy}}{{dx}} = a\frac{d}{{dx}}\left( {{e^x}} \right) + b\frac{d}{{dx}}\left( {{e^{ - x}}} \right) + \frac{d}{{dx}}\left( {{x^2}} \right)$

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

Again, differentiating both sides with respect to $x$,

we get:
$\frac{{{d^2}y}}{{d{x^2}}} = a{e^x} + b{e^{ - x}} + 2$

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

we get:
L.H.S.
$x\frac{{{d^2}y}}{{d{x^2}}} + 2\frac{{dy}}{{dx}} - xy + {x^2} - 2$
$= x\left( {a{e^x} + b{e^{ - x}} + 2} \right) + 2\left( {a{e^x} - b{e^{ - x}} + 2x} \right) - x\left( {a{e^x} + b{e^{ - x}} + {x^2}} \right) + {x^2} - 2$

$= \left( {ax{e^x} + bx{e^{ - x}} + 2x} \right) + \left( {2a{e^x} - 2b{e^{ - x}} + 4x} \right) - \left( {ax{e^x} + bx{e^{ - x}} + {x^3}} \right) + {x^2} - 2$

$= 2a{e^x} - 2b{e^{ - x}} + {x^2} + 6x - 2$
$\ne 0$

$\Rightarrow$ ${\rm{L}}.{\rm{H}}.{\rm{S}}. \ne {\rm{R}}.{\rm{H}}.{\rm{S}}$.

Hence, the given function is not a solution of the corresponding differential equation.
(ii) $y = {e^x}(a\cos x + b\sin x) = a{e^x}\cos x + b{e^x}\sin x$

Differentiating both sides with respect to ${\rm{x}}$,

we get:
$\frac{{dy}}{{dx}} = a \cdot \frac{d}{{dx}}\left( {{e^x}\cos x} \right) + b \cdot \frac{d}{{dx}}\left( {{e^x}\sin x} \right)$
$\Rightarrow$ $\frac{{dy}}{{dx}} = a\left( {{e^x}\cos x - {e^x}\sin x} \right) + b \cdot \left( {{e^x}\sin x + {e^x}\cos x} \right)$

$\Rightarrow$ $\frac{{dy}}{{dx}} = (a + b){e^x}\cos x + (b - a){e^x}\sin x$

Again, differentiating both sides with respect to ${\rm{x}}$,

we get:
$\frac{{{d^2}y}}{{d{x^2}}} = (a + b) \cdot \frac{d}{{dx}}\left( {{e^x}\cos x} \right) + (b - a)\frac{d}{{dx}}\left( {{e^x}\sin x} \right)$
$\Rightarrow$ $\frac{{{d^2}y}}{{d{x^2}}} = (a + b) \cdot \left[ {{e^x}\cos x - {e^x}\sin x} \right] + (b - a)\left[ {{e^x}\sin x + {e^x}\cos x} \right]$

$\Rightarrow$ $\frac{{{d^2}y}}{{d{x^2}}} = {e^x}[(a + b)(\cos x - \sin x) + (b - a)(\sin x + \cos x)]$

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

$\Rightarrow$ $\frac{{{d^2}y}}{{d{x^2}}} = \left[ {2{e^x}(b\cos x - a\sin x)} \right]$

Now, on substituting the values of$\frac{{{d^2}y}}{{d{x^2}}}$ and $\frac{{dy}}{{dx}}$ in the L.H.S. of the given differential equation,

we get:
$\frac{{{d^2}y}}{{d{x^2}}} + 2\frac{{dy}}{{dx}} + 2y$
$= 2{e^x}(b\cos x - a\sin x) - 2{e^x}[(a + b)\cos x + (b - a)\sin x] + 2{e^x}(a\cos x + b\sin x)$

$= {e^x}\left[ {\begin{array}{llllllllllllllllllll}{(2b\cos x - 2a\sin x) - (2a\cos x + 2b\cos x)}\\{ - (2b\sin x - 2a\sin x) + (2a\cos x + 2b\sin x)}\end{array}} \right]$

$= {e^x}[(2b - 2a - 2b + 2a)\cos x] + {e^x}[( - 2a - 2b + 2a + 2b)\sin x]$
$= 0$

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

(iii) $y = x\sin 3x$
Differentiating both sides with respect to ${\rm{x}}$,

we get:
$\frac{{dy}}{{dx}} = \frac{d}{{dx}}(x\sin 3x) = \sin 3x + x \cdot \cos 3x \cdot 3$
$\Rightarrow$ $\frac{{dy}}{{dx}} = \sin 3x + 3x\cos 3x$

Again, differentiating both sides with respect to ${\rm{x}}$,

we get:
$\frac{{{d^2}y}}{{d{x^2}}} = \frac{d}{{dx}}(\sin 3x) + 3\frac{d}{{dx}}(x\cos 3x)$
$\Rightarrow$ $\frac{{{d^2}y}}{{d{x^2}}} = 3\cos 3x + 3[\cos 3x + x( - \sin 3x) \cdot 3]$
$\Rightarrow$ $\frac{{{d^2}y}}{{d{x^2}}} = 6\cos 3x - 9x\sin 3x$

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

we get:
$\frac{{{d^2}y}}{{d{x^2}}} + 9y - 6\cos 3x$
$= (6 \cdot \cos 3x - 9x\sin 3x) + 9x\sin 3x - 6\cos 3x$
$= 0$

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

(iv) ${x^2} = 2{y^2}\log y$

Differentiating both sides with respect to $x$,

we get:
$2x = 2 \cdot \frac{d}{{dx}} = \left[ {{y^2}\log y} \right]$

$\Rightarrow$ $x = \left[ {2y \cdot \log y \cdot \frac{{dy}}{{dx}} + {y^2} \cdot \frac{1}{y} \cdot \frac{{dy}}{{dx}}} \right]$

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

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

we get:
$\left( {{x^2} + {y^2}} \right)\frac{{dy}}{{dx}} - xy$
$= \left( {2{y^2}\log y + {y^2}} \right) \cdot \frac{x}{{y(1 + 2\log y)}} - xy$

$= {y^2}(1 + 2\log y) \cdot \frac{x}{{y(1 + 2\log y)}} - xy$
$= xy - xy$
$= 0$

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