Analytical Solution of Ordinary Differential Equation

**NFS1** · August 7th 2012, 06:39 AM

Ive had a few attempts at trying to answer this question but have been unsuccessful. Can anyone guide me through the answer so I know how to do it.

Differential Equation ---> y'=(2/x)y+(x^2)(e^x)

Show that the analytical solution is given by ---> y(x) = x^2(e^x - e)

**JohnDMalcolm** · August 7th 2012, 06:56 AM

The question just asks to verify the given solution. Use this solution in the left- and right-hand sides of your differential equation and show that they are equal.

**Prove It** · August 7th 2012, 07:08 AM

Originally Posted by NFS1

Ive had a few attempts at trying to answer this question but have been unsuccessful. Can anyone guide me through the answer so I know how to do it.

Differential Equation ---> y'=(2/x)y+(x^2)(e^x)

Show that the analytical solution is given by ---> y(x) = x^2(e^x - e)

If you are going to solve the DE

$\displaystyle \begin{align*} \frac{dy}{dx} &= \left(\frac{2}{x}\right)y + x^2e^x \\ \frac{dy}{dx} - \left(\frac{2}{x}\right)y &= x^2e^x \end{align*}$

Now multiplying both sides by the Integrating Factor $\displaystyle \begin{align*} e^{\int{-\frac{2}{x}\,dx}} = e^{-2\ln{x}} = e^{\ln{\left(x^{-2}\right)}} = x^{-2} \end{align*}$ gives

$\displaystyle \begin{align*} x^{-2}\,\frac{dy}{dx} - 2x^{-3}y &= e^x \\ \frac{d}{dx}\left(x^{-2}y\right) &= e^x \\ x^{-2}y &= \int{e^x\,dx} \\ x^{-2}y &= e^x + C \\ y &= x^2\left(e^x + C\right) \end{align*}$

Obviously C = e is a possibility.

**HallsofIvy** · August 7th 2012, 08:03 AM

NFS1, do you understand that, as John D Malcolm said, this problem does NOT ask you to find the general solution? In this case it is fairly straightforward to solve, as Prove It showed, but you might have equations in the future that are much harder or impossible to solve in general- but it is easy to show that a given function satisfies the equation.

**NFS1** · August 7th 2012, 08:08 AM

Differential equation ----> xy'' + y' = x^2 can be written as d/dx(xy')=x^2 find analytical solution

So using steps Prove It has shown

> xy'=int(x^2)dx
> xy'=(x^3)/3
> y'=(x^2)/3
> y = (x^3)/9 as the analytical solution??

**NFS1** · August 7th 2012, 09:04 AM

As HallsofIvy and JohnD have said

> y=(x^3)/9
> y'=(x^2)/3
> y''=2x/3

substituting these into left and right hand side

>(2x^2)/3 + (x^2)/3 = x^2
>(3x^2)/3 = x^2
> x^2 = x^2

so they both equal and therefore my analytical solution is y=(x^3)/9 correct?

**Prove It** · August 7th 2012, 03:23 PM

Originally Posted by NFS1

Differential equation ----> xy'' + y' = x^2 can be written as d/dx(xy')=x^2 find analytical solution

So using steps Prove It has shown

> xy'=int(x^2)dx
> xy'=(x^3)/3
> y'=(x^2)/3
> y = (x^3)/9 as the analytical solution??

Where are your integration constants?

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