Solve this following

$\frac{\mathrm{dy}}{\mathrm{dx}}+\frac{\mathrm{y}}{\mathrm{x}}=\mathrm{bx}^{3}$

I.F. $=e^{\frac{1}{x} d x}=x$

So, solution of D.E. is given by

$y \cdot x=\int b \cdot x^{3} \cdot x d x+c$

$y=\frac{c}{x}+\frac{b x^{4}}{5}$

Passes through $(1,2)$

$2=\mathrm{c}+\frac{\mathrm{b}}{5}$          ..............(1)

$\int_{1}^{2} f(x) d x=\frac{62}{5}$

$\left[\mathrm{c} \ln \mathrm{x}+\frac{\mathrm{bx}^{5}}{25}\right]_{1}^{2}=\frac{62}{5}$

$\operatorname{cln} 2+\frac{31 \mathrm{~b}}{25}=\frac{62}{5}$  ...........(2)

By equation (1) \& (2)

$\mathrm{c}=0$ and $\mathrm{b}=10$