Solve the following differential equation dy dx -x ^2x=1 x - Vidyadip

Question

Solve the following differential equation
$\frac{\text{dy}}{\text{dx}}-\text{x}\sin^2\text{x}=\frac{1}{\text{x}\log\text{x}}$

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Answer

$\frac{\text{dy}}{\text{dx}}-\text{x}\sin^2\text{x}=\frac{1}{\text{x}\log\text{x}}$
$\frac{\text{dy}}{\text{dx}}=\frac{1}{\text{x}\log\text{x}}+\text{x}\sin^2\text{x}$
$\text{dy}=\Big(\frac{1}{\text{x}\log\text{x}}+\text{x}\sin^2\text{x}\Big)\text{dx}$
$\int\text{dy}=\int\frac{1}{\text{x}\log\text{x}}\text{dx}+\int\text{x}\sin^2\text{x dx}$
$\text{y}=\text{I}_1+\text{I}_2$
$\text{I}_1=\int\frac{1}{\text{x}\log\text{x}}\text{ dx}$
Let $\log\text{x}=\text{t}$
$\frac{1}{\text{x}}\ \text{dx}=\text{dt}$
$\text{I}_1=\int\frac{\text{dt}}{\text{t}}$
$=\log|\text{t}|+\text{C}_1$
$\text{I}_1=\log|\log\text{x}|+\text{C}_1$
$\text{I}_2=\int\text{x}\sin^2\text{x dx}$
$=\int\text{x}\frac{(1-\cos2\text{x})}{2}\ \text{dx}$
$=\frac{1}{2}\int(\text{x}-\text{x}\cos2\text{x})\text{dx}$
$=\frac{1}{2}\int\text{x dx}-\frac{1}{2}\int\text{x}\cos2\text{x dx}$
$=\frac{1}{2}\Big(\frac{\text{x}^2}{2}\Big)-\frac{1}{2}[\text{x}\int\cos2\text{x dx}-\int(1\times\int\cos2\text{x dx})\text{dx}]+\text{C}_2$
$=\frac{\text{x}^2}{4}-\frac{1}{2}\Big[\frac{\text{x}\sin\text{x}}{2}+\frac{\cos2\text{x}}{4}\Big]+\text{C}_2$
$\text{I}_2=\frac{\text{x}^2}{4}-\frac{\text{x}\sin2\text{x}}{4}-\frac{\cos2\text{x}}{8}+\text{C}_2$
Put the values of $I_1$ and $I_2$ in equation $(1)$
$\text{y}=\text{I}_1+\text{I}_2$
$\text{y}=\log|\log\text{x}|+\frac{\text{x}^2}{4}-\frac{\text{x}\sin2\text{x}}{4}-\frac{\cos2\text{x}}{8}+\text{C}$ as $\text{ C}_1+\text{C}_2=\text{C}$

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