Question

solve the differential equation dy/dx=x2y+y​

Answer 1

We're given to solve the differential equation,

$\longrightarrow\dfrac{dy}{dx}=x^2y+y$

or,

$\longrightarrow\dfrac{dy}{dx}=\left(x^2+1\right)y$

$\longrightarrow\dfrac{1}{y}\,dy=\left(x^2+1\right)\ dx$

Integrating,

$\displaystyle\longrightarrow\int\dfrac{1}{y}\,dy=\int\left(x^2+1\right)\ dx$

$\displaystyle\longrightarrow\log|y|-\log|C|=\dfrac{x^3}{3}+x$

$\displaystyle\longrightarrow\log\left|\dfrac{y}{C}\right|=\dfrac{x^3}{3}+x$

Taking antilog,

$\displaystyle\longrightarrow\dfrac{y}{C}=e^{\frac{x^3}{3}+x}$

$\displaystyle\longrightarrow\underline{\underline{y=Ce^{\frac{x^3}{3}+x}}}$

This is the solution to the differential equation, where C is an arbitrary constant.

Answer 2

$\large\underline\purple{\bold{Solution :- }}$

$\rm :\implies\:\dfrac{dy}{dx} = {x}^{2} y + y$

$\rm :\implies\:\dfrac{dy}{dx} = y( {x}^{2} + 1)$

$\rm :\implies\:1/y \: dy \: = ( {x}^{2} + 1)dx$

So, on integrating both sides, we get

$\rm :\implies\: \int \: 1/y \: dy \: = \int \: ( {x}^{2} + 1)dx$

$\rm :\implies\:logy = \dfrac{ {x}^{3} }{3} + x + c$