GATE Civil Engineering (CE) 2019 Shift 1 Solved Paper

${\mathit{x}}^2\frac{\mathit{d}\mathit{y}}{\mathit{d}{\mathit{x}}^2}-2\mathit{x}\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}+2\mathit{y}=0$

Assume $\mathit{x}={\mathit{e}}^{\mathit{t}}$
$\Rightarrow \frac{\mathit{d}\mathit{x}}{\mathit{d}\mathit{t}}=6^{\mathit{t}}=\mathit{x}$
$\Rightarrow \mathit{d}\mathit{x}=\mathit{x}\mathit{d}\mathit{t}$
Then $\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}=\frac{\mathit{d}\mathit{y}}{\mathit{x}\mathit{d}\mathit{t}}$

$\Rightarrow {\mathit{x}}^{\frac{\mathit{d}}{\mathit{d}\mathit{t}}}\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{t}}=\mathit{D}\left(\mathit{D}-1\right)\mathit{y}$
Putting this into ordinary differential equation
$\mathit{D}\left(\mathit{D}-1\right)\mathit{y}-2\mathit{D}\mathit{y}+2\mathit{y}=0$
$\left[{\mathit{D}}^2-3\mathit{D}+2\right]=0$
Auxiliary equation is
${\mathit{m}}^2-3\mathit{m}+2=0$
$\left(\mathit{m}-1\right)\left(\mathit{m}-2\right)=0$
i.e.
$\therefore$ solution of equation
$\mathit{y}={\mathit{c}}_1{\mathit{e}}^{\mathit{t}}+{\mathit{c}}_2{\mathit{e}}^{2\mathit{t}}$
$\Rightarrow$ Putting et as $\mathit{x}$
$\Rightarrow \mathit{y}={\mathit{c}}_1\mathit{x}+{\mathit{c}}_2{\mathit{X}}^2$
$\Rightarrow \mathit{y}\left(1\right)=0$
$\Rightarrow {\mathit{c}}_1+{\mathit{c}}_2=0$.....(i)

$\Rightarrow 2{\mathit{c}}_1+4{\mathit{c}}_2=2$
$\Rightarrow {\mathit{c}}_1+2{\mathit{c}}_2=1$.....(ii)

$\Rightarrow {\mathit{c}}_2=1$
${\mathit{c}}_1=-1$
$\Rightarrow \mathit{y}=-\mathit{x}+{\mathit{x}}^2$
Then $\mathit{y}\left(3\right)=-3+3^2=6$