gaby131o

2022-09-21

What is the general solution of the differential equation $y\prime -2xy={x}^{3}$?

Katelyn Ryan

Beginner2022-09-22Added 11 answers

We have:

$y\prime -2xy={x}^{3}$ ..... [1]

This is 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 can readily generate an integrating factor when we have an equation of this form, given by;

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

$=\mathrm{exp}(\int -2xdx)$

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

$={e}^{-{x}^{2}}$

And if we multiply the DE [1] by this Integrating Factor, I, we will have a perfect product differential;

$e}^{-{x}^{2}}y\prime -2x{e}^{-{x}^{2}}y={x}^{3}{e}^{-{x}^{2}$

$\therefore \frac{d}{dx}\left({e}^{-{x}^{2}}y\right)={x}^{3}{e}^{-{x}^{2}}$

We can now integrate to get:

${e}^{-{x}^{2}}y=\int {x}^{3}{e}^{-{x}^{2}}dx+C$

The RHS integral can be evaluated by an application of Integration By Parts (omitted) which gives us:

$\int {x}^{3}{e}^{-{x}^{2}}dx=-\frac{1}{2}({x}^{2}+1){e}^{-{x}^{2}}$

So we have:

${e}^{-{x}^{2}}y=-\frac{1}{2}({x}^{2}+1){e}^{-{x}^{2}}+C$

Using the initial condition y(0)=2 we can evaluate the constant C

$2{e}^{0}=-\frac{1}{2}(0+1){e}^{0}+C\Rightarrow C=\frac{5}{2}$

Leading to the Particular Solution:

$e}^{-{x}^{2}}y=-\frac{1}{2}({x}^{2}+1){e}^{-{x}^{2}}+\frac{5}{2$

$\therefore y=-\frac{1}{2}({x}^{2}+1){e}^{-{x}^{2}}+\frac{5}{2{e}^{-{x}^{2}}}$

$=-\frac{1}{2}({x}^{2}+1)+\frac{5}{2}{e}^{{x}^{2}}$

$=\frac{5}{2}{e}^{{x}^{2}}-\frac{1}{2}{x}^{2}-\frac{1}{2}$

$y\prime -2xy={x}^{3}$ ..... [1]

This is 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 can readily generate an integrating factor when we have an equation of this form, given by;

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

$=\mathrm{exp}(\int -2xdx)$

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

$={e}^{-{x}^{2}}$

And if we multiply the DE [1] by this Integrating Factor, I, we will have a perfect product differential;

$e}^{-{x}^{2}}y\prime -2x{e}^{-{x}^{2}}y={x}^{3}{e}^{-{x}^{2}$

$\therefore \frac{d}{dx}\left({e}^{-{x}^{2}}y\right)={x}^{3}{e}^{-{x}^{2}}$

We can now integrate to get:

${e}^{-{x}^{2}}y=\int {x}^{3}{e}^{-{x}^{2}}dx+C$

The RHS integral can be evaluated by an application of Integration By Parts (omitted) which gives us:

$\int {x}^{3}{e}^{-{x}^{2}}dx=-\frac{1}{2}({x}^{2}+1){e}^{-{x}^{2}}$

So we have:

${e}^{-{x}^{2}}y=-\frac{1}{2}({x}^{2}+1){e}^{-{x}^{2}}+C$

Using the initial condition y(0)=2 we can evaluate the constant C

$2{e}^{0}=-\frac{1}{2}(0+1){e}^{0}+C\Rightarrow C=\frac{5}{2}$

Leading to the Particular Solution:

$e}^{-{x}^{2}}y=-\frac{1}{2}({x}^{2}+1){e}^{-{x}^{2}}+\frac{5}{2$

$\therefore y=-\frac{1}{2}({x}^{2}+1){e}^{-{x}^{2}}+\frac{5}{2{e}^{-{x}^{2}}}$

$=-\frac{1}{2}({x}^{2}+1)+\frac{5}{2}{e}^{{x}^{2}}$

$=\frac{5}{2}{e}^{{x}^{2}}-\frac{1}{2}{x}^{2}-\frac{1}{2}$

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