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The general solution of the differential equation $x^{2} d y-2 x y d x=x^{4} \cos x d x$ is
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Verified Answer
The correct answer is:
$y=x^{2} \sin x+c x^{2}$
We have,
$\begin{aligned}
x^{2} d y-2 x y d x &=x^{4} \cos x d x \\
\Rightarrow \quad \quad \frac{d y}{d x}-\frac{2}{x} y &=x^{2} \cos x
\end{aligned}$
Comparing with $\frac{d y}{d x}+P y=Q$
Where $P=-\frac{2}{x}$ and $Q=x^{2} \cos x$
$\therefore \quad \text { IF }=e^{\int \frac{-2}{x} d x}=e^{-2 \log x}=e^{\log \left(\frac{1}{x^{2}}\right)}=\frac{1}{x^{2}}$
$\therefore$ The solution of the given differential
$y \cdot \operatorname{IF}=\int(Q \times \operatorname{IF}) d x+c$
$\Rightarrow y \times \frac{1}{x^{2}}=\int\left(x^{2} \cos x\right)\left(\frac{1}{x^{2}}\right) d x+c$
$\Rightarrow \quad \frac{y}{x^{2}}=\int \cos x+c$
$\Rightarrow \quad \frac{y}{x^{2}}=\sin x+c$
$\Rightarrow \quad y=x^{2} \sin x+c x^{2}$
$\begin{aligned}
x^{2} d y-2 x y d x &=x^{4} \cos x d x \\
\Rightarrow \quad \quad \frac{d y}{d x}-\frac{2}{x} y &=x^{2} \cos x
\end{aligned}$
Comparing with $\frac{d y}{d x}+P y=Q$
Where $P=-\frac{2}{x}$ and $Q=x^{2} \cos x$
$\therefore \quad \text { IF }=e^{\int \frac{-2}{x} d x}=e^{-2 \log x}=e^{\log \left(\frac{1}{x^{2}}\right)}=\frac{1}{x^{2}}$
$\therefore$ The solution of the given differential
$y \cdot \operatorname{IF}=\int(Q \times \operatorname{IF}) d x+c$
$\Rightarrow y \times \frac{1}{x^{2}}=\int\left(x^{2} \cos x\right)\left(\frac{1}{x^{2}}\right) d x+c$
$\Rightarrow \quad \frac{y}{x^{2}}=\int \cos x+c$
$\Rightarrow \quad \frac{y}{x^{2}}=\sin x+c$
$\Rightarrow \quad y=x^{2} \sin x+c x^{2}$
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