Question
Solve $\text{y}\text{dx}-\text{x}\text{dy}=\text{x}^2\text{y}\text{dx}.$
✓
Answer
Given that, $\text{y}\text{dx}-\text{x}\text{dy}=\text{x}^2\text{y}\text{dx}$
$\Rightarrow\frac{1}{\text{x}^2}-\frac{1}{\text{xy}}.\frac{\text{dy}}{\text{dx}}=1$ [dividing throughout by x2ydx]
$\Rightarrow-\frac{1}{\text{xy}}.\frac{\text{dy}}{\text{dx}}+\frac{1}{\text{x}^2}-1=0$
$\Rightarrow\frac{\text{dy}}{\text{dx}}-\frac{\text{xy}}{\text{x}^2}+\text{xy}=0$
$\Rightarrow\frac{\text{dy}}{\text{dx}}-\frac{\text{y}}{\text{x}}+\text{xy}=0$
$\Rightarrow\frac{\text{dy}}{\text{dx}}+\Big(\text{x}-\frac{1}{\text{x}}\Big)\text{y}=0$
Which is linear differential equation.
On comparing it with $\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q},$ we get
$\text{P}=\Big(\text{x}-\frac{1}{\text{x}}\Big),\text{Q}=0$
$\text{I.f}=\text{e}^{\int\text{P}\text{dx}}$
$=\text{e}^{\int\big(\text{x}-\frac{1}{\text{x}}\big)}\text{dx}$
$=\text{e}^{\frac{\text{x}^2}{2}-\log\text{x}}$
$=\text{e}^{\frac{\text{x}^2}{2}}.\text{e}^{-\log\text{x}}$
$=\frac{1}{\text{x}}\text{e}^{\frac{\text{x}^2}{2}}$
The general solution is,
$\text{y}.\frac{1}{\text{x}}\text{e}^\frac{\text{x}^2}{2}=\int0.\frac{1}{\text{x}}\text{e}^{\frac{\text{x}^2}{2}}\text{dx}+\text{C}$
$\Rightarrow\text{y}.\frac{1}{\text{x}}\text{e}^{\frac{\text{x}^2}{2}}=\text{C}$
$\Rightarrow\text{y}=\text{C}\text{x}\text{e}^{\frac{-\text{x}^2}{2}}$
$\Rightarrow\frac{1}{\text{x}^2}-\frac{1}{\text{xy}}.\frac{\text{dy}}{\text{dx}}=1$ [dividing throughout by x2ydx]
$\Rightarrow-\frac{1}{\text{xy}}.\frac{\text{dy}}{\text{dx}}+\frac{1}{\text{x}^2}-1=0$
$\Rightarrow\frac{\text{dy}}{\text{dx}}-\frac{\text{xy}}{\text{x}^2}+\text{xy}=0$
$\Rightarrow\frac{\text{dy}}{\text{dx}}-\frac{\text{y}}{\text{x}}+\text{xy}=0$
$\Rightarrow\frac{\text{dy}}{\text{dx}}+\Big(\text{x}-\frac{1}{\text{x}}\Big)\text{y}=0$
Which is linear differential equation.
On comparing it with $\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q},$ we get
$\text{P}=\Big(\text{x}-\frac{1}{\text{x}}\Big),\text{Q}=0$
$\text{I.f}=\text{e}^{\int\text{P}\text{dx}}$
$=\text{e}^{\int\big(\text{x}-\frac{1}{\text{x}}\big)}\text{dx}$
$=\text{e}^{\frac{\text{x}^2}{2}-\log\text{x}}$
$=\text{e}^{\frac{\text{x}^2}{2}}.\text{e}^{-\log\text{x}}$
$=\frac{1}{\text{x}}\text{e}^{\frac{\text{x}^2}{2}}$
The general solution is,
$\text{y}.\frac{1}{\text{x}}\text{e}^\frac{\text{x}^2}{2}=\int0.\frac{1}{\text{x}}\text{e}^{\frac{\text{x}^2}{2}}\text{dx}+\text{C}$
$\Rightarrow\text{y}.\frac{1}{\text{x}}\text{e}^{\frac{\text{x}^2}{2}}=\text{C}$
$\Rightarrow\text{y}=\text{C}\text{x}\text{e}^{\frac{-\text{x}^2}{2}}$
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