Angelique Horne

2022-05-13

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

Giancarlo Shah

Beginner2022-05-14Added 12 answers

We can use an integrating factor when we have a First Order Linear non-homogeneous Ordinary Differential Equation of the form;

$\frac{dy}{dx}+P(x)y=Q(x)$

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;

$I={e}^{\int P(x)dx}$

$=\mathrm{exp}(\int {x}^{2}dx)$

$=\mathrm{exp}(\frac{1}{3}{x}^{3})$

=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;

${y}^{\prime}{e}^{\frac{1}{3}{x}^{3}}+{x}^{2}{e}^{\frac{1}{3}{x}^{3}}y={x}^{2}{e}^{\frac{1}{3}{x}^{3}}$

$\therefore \frac{d}{dx}(y{e}^{\frac{1}{3}{x}^{3}})dx+C$

Leading to the explicit General Solution:

$y={e}^{-\frac{1}{3}{x}^{3}}\{{e}^{\frac{1}{3}{x}^{3}}+C\}$

$=1+C{e}^{-\frac{1}{3}{x}^{3}}$

$\frac{dy}{dx}+P(x)y=Q(x)$

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;

$I={e}^{\int P(x)dx}$

$=\mathrm{exp}(\int {x}^{2}dx)$

$=\mathrm{exp}(\frac{1}{3}{x}^{3})$

=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;

${y}^{\prime}{e}^{\frac{1}{3}{x}^{3}}+{x}^{2}{e}^{\frac{1}{3}{x}^{3}}y={x}^{2}{e}^{\frac{1}{3}{x}^{3}}$

$\therefore \frac{d}{dx}(y{e}^{\frac{1}{3}{x}^{3}})dx+C$

Leading to the explicit General Solution:

$y={e}^{-\frac{1}{3}{x}^{3}}\{{e}^{\frac{1}{3}{x}^{3}}+C\}$

$=1+C{e}^{-\frac{1}{3}{x}^{3}}$

vilitatelp014

Beginner2022-05-15Added 6 answers

${y}^{\prime}+{x}^{2}y={x}^{2}$

This is a separable ODE

$\frac{dy}{dx}={x}^{2}-{x}^{2}y={x}^{2}(1-y)$

$\Rightarrow \int \frac{1}{1-y}dy=\int {x}^{2}dx$

$\Rightarrow -\mathrm{ln}|1-y|=\frac{1}{3}{x}^{3}+c$

$1-y=B{e}^{\frac{1}{3}{x}^{3}}$ where $B={e}^{-c}$

$y=1+A{e}^{\frac{1}{3}{x}^{3}}$ where A=-B

This is a separable ODE

$\frac{dy}{dx}={x}^{2}-{x}^{2}y={x}^{2}(1-y)$

$\Rightarrow \int \frac{1}{1-y}dy=\int {x}^{2}dx$

$\Rightarrow -\mathrm{ln}|1-y|=\frac{1}{3}{x}^{3}+c$

$1-y=B{e}^{\frac{1}{3}{x}^{3}}$ where $B={e}^{-c}$

$y=1+A{e}^{\frac{1}{3}{x}^{3}}$ where A=-B

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