Solve the following differential equation: y^2 dx dy +x-1 y =0 - Vidyadip

Question

Solve the following differential equation:
$\text{y}^2\frac{\text{dx}}{\text{dy}}+\text{x}-\frac{1}{\text{y}}=0$

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Answer

Here, $\text{y}^2\frac{\text{dx}}{\text{dy}}+\text{x}-\frac{1}{\text{y}}=0$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}+\frac{\text{x}}{\text{y}^2}=\frac{1}{\text{y}^3}$
It is a linear differential equation. Comparing the equation with,
$\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q}$
$\text{P}=\frac{1}{\text{y}^2},\text{Q}=\frac{1}{\text{y}^3}$
I.F. $=\text{e}^{\int\text{Pdy}}$
$=\text{e}^{\int\frac{1}{\text{y}^2}\text{dy}}$
$=\text{e}^{-\frac{1}{\text{y}}}$
Solution of the equation is given by,
$\text{x}\times(\text{I.F.})=\int\text{Q}\times(\text{I.F.})\text{dy + C}$
$\text{x}\Big(\text{e}^{-\frac{1}{\text{y}}}\Big)=\int\frac{1}{\text{y}^3}\Big(\text{e}^{-\frac{1}{\text{y}}}\Big)\text{dy + C}$
Let $\text{e}^{-\frac{1}{\text{y}}}=\text{t}$
$\Rightarrow\frac{1}{\text{y}}=-\log\text{t}$
$\text{e}^{-\frac{1}{\text{y}}}\times\frac{1}{\text{y}^2}\text{dy = dt}$
$\text{x (t)}=\int\frac{1}{\text{y}}\text{dt + C}$
$=-\log+\text{dt + C}$
$=-\Big[\log\text{t}\times\int1\times\text{dt}-\int\Big(\frac{1}{\text{t}}\int1\times\text{dt}\Big)\text{dt}\Big]+\text{C}$
$=-\Big[\text{t}\log\text{t}-\int\frac{\text{t}}{\text{t}}\text{dt}\Big]+\text{C}$
$\text{x (t)}=-\text{t}\log\text{t + t + C}$
$\text{x (t)}=-\text{t}[\log\text{t}-1]+\text{C}$
$\text{x}=-\Big[-\frac{1}{\text{y}}-1\Big]\text{Ce}^{\frac{1}{\text{y}}}$
$\text{x}=\frac{1}{\text{y}}+1+\text{Ce}^{\frac{1}{\text{y}}}$
$\text{x}=\Big(\frac{1+\text{y}}{\text{y}}\Big)+\text{Ce}^{\frac{1}{\text{y}}}$

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