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

$\mathrm{I}.\mathrm{F}.=e^{\int \frac{1}{ x}dx}=x$

So, solution of $\mathrm{D}.\mathrm{E}.$ is given by

$y·x=\int b·x^3·xdx+c$

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

Passes through $\left(1, 2\right)$

$2=c+\frac{b}{5} \dots \left(1\right)$

${\int }_1^{2}f\left(x\right)dx=\frac{62}{5}$

${\left[c\mathit{ }\ln x+\frac{b{x}^5}{25}\right]}_1^2=\frac{62}{5}$

$\mathit{c}\mathit{ }\ln \mathit{ }2+\frac{31 b}{25}=\frac{62}{5} \dots \left(2\right)$

By equation $\left(1\right)$ and $\left(2\right)$

$c=0$ and $b=10$