GATE Civil Engineering (CE) 2020 Shift 2 Solved Paper

$\frac{\partial \mathit{u}}{\partial \mathit{y}}=\mathit{x}{\mathit{y}}^{\prime }$

Given


$\mathit{y}\prime -\mathit{y}\mathit{k}=0$ ......(1)
.....(2)
Take equation (1)
Auxiliary equation is
$\mathit{D}-\mathit{k}=0$
$\mathit{D}=\mathit{k}$
So the solution is
$\mathit{y}={\mathit{C}}_0{\mathit{e}}^{\mathit{k}\mathit{y}}$
Take equation (2)
${\mathit{x}}^{\prime \prime }-\mathit{x}\mathit{k}=0$
Auxiliary equation is
${\mathit{D}}^2-\mathit{k}=0$
$\mathit{D}=\pm \sqrt{\mathit{k}}$
So the solution is
$\mathit{X}={\mathit{C}}_1{\mathit{e}}^{\sqrt{\mathit{k}\mathit{x}}}+{\mathit{C}}_2{\mathit{e}}^{-\sqrt{\mathit{k}\mathit{x}}}$
But $\mathit{u}=\mathit{x}\mathit{y}$
$=\left({\mathit{C}}_1{\mathit{E}}^{\mathit{J}\mathit{k}\mathit{x}}+{\mathit{C}}_2{\mathit{e}}^{-\sqrt{\mathit{k}}\mathit{x}}\right){\mathit{C}}_0{\mathit{e}}^{\mathit{k}\mathit{y}}$
$={\mathit{C}}_1{\mathit{C}}_0{\mathit{e}}^{\sqrt{\mathit{k}}\mathit{x}\mathit{k}\mathit{y}}+{\mathit{C}}_2{\mathit{C}}_0{\mathit{e}}^{-\sqrt{\mathit{k}}\mathit{k}\mathit{x}\mathit{y}}$
$={\mathit{k}}_1{\mathit{e}}^{{\mathit{k}}^{3∕2}\mathit{x}\mathit{y}}+{\mathit{k}}_2{\mathit{e}}^{-{\mathit{k}}^{3∕2}\mathit{x}\mathit{y}}$
Hence option (A).