0
$\begingroup$

I want to solve the ODE $y''+4y=x^2+3e^x$

I already found the complemenetary homogenous solutions: $y_1=\cos (2x)$ and $y_2=\sin (2x)$ and also found the wronskian: $|W|=2$

Now, according to http://tutorial.math.lamar.edu/Classes/DE/VariationofParameters.aspx the variation of parameters method states that the particular solution is:

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

This isn't the most pleasant integral I've seen. Far from it, it doesn't seem feasible to calculate it. Is there another method to find the particular solution?

$\endgroup$
  • $\begingroup$ Those integrals can be computed by parts, with multiple rounds. $\endgroup$ – user170231 Feb 4 '15 at 21:40
2
$\begingroup$

Yes there is another method which is known as the undetermined coefficients. According to it we assume the particular solution to have the form

$$ y_p = A+Bx+cx^2 + De^x \longrightarrow (1) $$

where constants $A,B,C,D$ need to be determined by substituting $ y_p, y'_p, y''_p $ in the ode.

$\endgroup$
0
$\begingroup$

The method of undetermined coefficients makes things much simpler.

$\endgroup$
0
$\begingroup$

It is not difficult to check that: $$ f(x) = \frac{x^2}{4}-\frac{1}{8}+\frac{3}{5}e^x $$ is a solution, so by setting $y(x) = g(x)+f(x)$ we have that $g$ satisfies the homogeneous ODE: $$ g''+4g = 0,$$ for which you already know the solutions.

$\endgroup$
0
$\begingroup$

I am a fan of the Laplace transform technique: $$\mathcal{L}\left\{y''+4y=x^2+3e^x \right\} \\ \implies s^2Y(s)-y'(0)-sy(0)+4Y(s)=\frac{2}{s^3}+\frac{3}{s-1} \\ \implies Y(s)(s^2+4) = \frac{2}{s^3}+\frac{3}{s-1}+y'(0)+sy(0) \\ \implies Y(s) = \frac{2}{s^3(s^2+4)}+\frac{3}{(s-1)(s^2+4)}+\frac{y'(0)}{s^2+4}+\frac{sy(0)}{s^2+4}$$ Then use partial fraction decomposition on $\frac{2}{s^3(s^2+4)}$ and $\frac{3}{(s-1)(s^2+4)}$ to get $$\frac{2}{s^3(s^2+4)}=\frac{1}{2s^3}+\frac{s}{8(s^2+4)}-\frac{1}{8s} \\ \frac{3}{(s-1)(s^2+4)}= \frac{3}{5(s-1)}-\frac{3s}{5(s^2+4)}-\frac{3}{5(s^2+4)}$$ Hence $$Y(s) = \frac{1}{2s^3}+\frac{s}{8(s^2+4)}-\frac{1}{8s}+\frac{3}{5(s-1)}-\frac{3s}{5(s^2+4)}-\frac{3}{5(s^2+4)}+\frac{y'(0)}{s^2+4}+\frac{sy(0)}{s^2+4}$$ and undoing the transform should reveal $$y(x) = \frac{x^2}{4}+\frac{1}{8}\cos(2x)-\frac{1}{8}+\frac{3}{5}e^x-\frac{3}{5}\cos(2x)-\frac{3}{10}\sin(2x)+\frac{y'(0)}{2}\sin(2x)+y(0)\cos(2x) \\ = \frac{x^2}{4}-\frac{1}{8}+\cos(2x)\left(y(0)-\frac{19}{40}\right)+\sin(2x)\left(y'(0)-\frac{3}{10} \right)+\frac{3}{5}e^x$$ This is the solution for whatever initial conditions $y(0), \space y'(0)$ you want. While the method is algebra intensive, you don't have to integrate anything.

$\endgroup$
0
$\begingroup$

$$\begin{align*}\int\frac{\sin2x(x^2+3e^x)}{2}\,dx&=\frac{1}{2}\left(\int x^2\sin2x\,dx+3\int e^x\sin2x\right)\end{align*}$$ Integrating by parts, you can find that $$\begin{align*}\int x^2\sin2x\,dx&=-\frac{1}{2}x^2\cos2x+\int x\cos2x\,dx\\ &=-\frac{1}{2}x^2\cos2x+\frac{1}{2}x\sin2x-\frac{1}{2}\int\sin2x\,dx\\ &=-\frac{1}{2}x^2\cos2x+\frac{1}{2}x\sin2x+\frac{1}{4}\cos2x+C_1\end{align*}$$ Similarly, $$\begin{align*} \int e^x\sin2x&=-\frac{1}{2}e^x\cos2x+\frac{1}{2}\int e^x\cos2x\,dx\\ &=-\frac{1}{2}e^x\cos2x+\frac{1}{2}\left(\frac{1}{2}e^x\sin2x-\frac{1}{2}\int e^x\sin2x\,dx\right)\\ &=-\frac{1}{2}e^x\cos2x+\frac{1}{4}e^x\sin2x-\frac{1}{4}\int e^x\sin2x\,dx\\ \frac{5}{4}\int e^x\sin2x\,dx&=-\frac{1}{2}e^x\cos2x+\frac{1}{4}e^x\sin2x\\ \int e^x\sin2x\,dx&=-\frac{2}{5}e^x\cos2x+\frac{1}{5}e^x\sin2x+C_2\end{align*}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.