The general solution of the differential equation \(x^{2} d y-2 x y d x=x^{4} \cos x d x\) is

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
\(y \cdot \operatorname{IF}=\int(Q \times \operatorname{IF}) d x+c\)
\(\Rightarrow y \times \frac{1}{x^{2}}=\int\left(x^{2} \cos x\right)\left(\frac{1}{x^{2}}\right) d x+c\)
\(\Rightarrow \quad \frac{y}{x^{2}}=\int \cos x+c\)
\(\Rightarrow \quad \frac{y}{x^{2}}=\sin x+c\)
\(\Rightarrow \quad y=x^{2} \sin x+c x^{2}\)