$\text{ The solution of }\frac{dy}{dx}+\frac{3}{x}y=\frac{1}{x^2}\text{ , given that }y=2,x=1\text{ is }2x^{3}y=x^3+\lambda \text{ then }\lambda \text{ is }$

$\begin{array}{l}I.F{\mathrm{e}}^{\int \frac{3dx}{\mathrm{x}}}={\mathrm{e}}^{3\log x}={\mathrm{x}}^3\\ \text{ Solution of the differential equation is }{\mathrm{yx}}^3=\int \frac{{\mathrm{x}}^3}{{\mathrm{x}}^2}\mathrm{dx}=\frac{{\mathrm{x}}^2}{2}+\mathrm{c}\\ \text{ Pass }\left(1.2\right)\Rightarrow c=3/2\\ \Rightarrow {\mathrm{yx}}^3=\frac{{\mathrm{x}}^2}{2}+\frac{3}{2}\\ \Rightarrow 2x^{3}y=x^2+3\\ \Rightarrow \lambda =3\text{ . }\end{array}$