Angelique Horne

2022-05-13

How can i solve this differencial equation?: ${y}^{\prime }+{x}^{2}y={x}^{2}$

Giancarlo Shah

We can use an integrating factor when we have a First Order Linear non-homogeneous Ordinary Differential Equation of the form;
$\frac{dy}{dx}+P\left(x\right)y=Q\left(x\right)$
We have:
${y}^{\prime }={x}^{2}y={x}^{2}$...[1]
This is a First Order Ordinary Differential Equation in Standard Form. So we compute and integrating factor, I, using;
$I={e}^{\int P\left(x\right)dx}$
$=\mathrm{exp}\left(\int {x}^{2}dx\right)$
$=\mathrm{exp}\left(\frac{1}{3}{x}^{3}\right)$
=e^{\fra$={e}^{\frac{1}{3}{x}^{3}}$
And if we multiply the DE [1] by this Integrating Factor, I, we will have a perfect product differential;
${y}^{\prime }{e}^{\frac{1}{3}{x}^{3}}+{x}^{2}{e}^{\frac{1}{3}{x}^{3}}y={x}^{2}{e}^{\frac{1}{3}{x}^{3}}$
$\therefore \frac{d}{dx}\left(y{e}^{\frac{1}{3}{x}^{3}}\right)dx+C$
Leading to the explicit General Solution:
$y={e}^{-\frac{1}{3}{x}^{3}}\left\{{e}^{\frac{1}{3}{x}^{3}}+C\right\}$
$=1+C{e}^{-\frac{1}{3}{x}^{3}}$

vilitatelp014

${y}^{\prime }+{x}^{2}y={x}^{2}$
This is a separable ODE
$\frac{dy}{dx}={x}^{2}-{x}^{2}y={x}^{2}\left(1-y\right)$
$⇒\int \frac{1}{1-y}dy=\int {x}^{2}dx$
$⇒-\mathrm{ln}|1-y|=\frac{1}{3}{x}^{3}+c$
$1-y=B{e}^{\frac{1}{3}{x}^{3}}$ where $B={e}^{-c}$
$y=1+A{e}^{\frac{1}{3}{x}^{3}}$ where A=-B

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