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  • Solve the differential equation Differential Equations Miscellaneous Exercise 12.

Q12.    Solve the differential equation \left[\frac{e^{-2\sqrt x}}{\sqrt x} - \frac{y}{\sqrt x} \right ]\frac{dx}{dy} = 1\; \ (x\neq 0).

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Given,
${\left[\frac{e^{-2 \sqrt{x}}}{\sqrt{x}}-\frac{y}{\sqrt{x}}\right] \frac{d x}{d y}=1}$
$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{e}^{-2 \sqrt{\mathrm{x}}}}{\sqrt{\mathrm{x}}}-\frac{\mathrm{y}}{\sqrt{\mathrm{x}}}$
$\Longrightarrow \frac{\mathrm{dy}}{\mathrm{dx}}+\frac{\mathrm{y}}{\sqrt{\mathrm{x}}}=\frac{\mathrm{e}^{-2 \sqrt{\mathrm{x}}}}{\sqrt{\mathrm{x}}}$
This is equation is in the form of $\frac{d y}{d x}+p y=Q$
$p=\frac{1}{\sqrt{x}} \text { and } Q=\frac{e^{-2 \sqrt{x}}}{\sqrt{x}}$
Now, I.F. $=\mathrm{e}^{\int \mathrm{pdx}}=\mathrm{e}^{\int \frac{1}{\sqrt{\mathrm{x}}} \mathrm{dx}}=\mathrm{e}^{2 \sqrt{\mathrm{x}}}$
We know that the solution of the given differential equation is:
$y(I \cdot F .)=\int(Q \cdot F \cdot) d x+C$
$\Rightarrow \mathrm{ye}^{2 \sqrt{\mathrm{x}}}=\int\left(\frac{\mathrm{e}^{-2 \sqrt{\mathrm{x}}}}{\sqrt{\mathrm{x}}} \times \mathrm{e}^{2 \sqrt{\mathrm{x}}}\right) \mathrm{dx}+C$
$\Rightarrow \mathrm{ye}^{2 \sqrt{\mathrm{x}}}=\int \frac{1}{\sqrt{\mathrm{x}}} \mathrm{dx}+\mathrm{C}$
$\Rightarrow \mathrm{ye}^{2 \sqrt{x}}=2 \sqrt{\mathrm{x}}+C$

Posted by

HARSH KANKARIA

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