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I need some help in finding the solution to this second-order non-homogeneous DE.

I know how to find the solution of the reduced equation $$y''+2y'+y=0.$$

The characteristic equation $r^{2}+2r+1=0$ yields real and repeated roots $r=-1$. So the complementary solution, $$y_{c}= c_{1}e^{-t} + c_{2}te^{-t}$$.

Now what I am struggling with is making educated guesses about coming up with a particular solution. So any suggestions on that would be helpful.

EDIT1: I haven't learned about the method of variation of parameters yet and wanted to do this with the method of undetermined coefficients. Please feel free to post the solution by using any of the techniques so I can learn from it.

EDIT2: Our fundamental pair of solutions for the corresponding homogeneous is $y_{1}=e^{-t}$, and $y_{2}=t e^{-t}$.

We know that our particular solution will be of the form, $y_{p} = u_{1} y_{1} + u_{2} y_{2}$.

The determinant of the Wronskian, $W = e^{-2t}$, with $W_{1} = -t^{-1} e^{-2t}$, and $W_{2} = t^{-1} e^{-2t}$.

So $u'_{1} = \frac{W_{1}}{W} = -\frac{1}{t}$, and $u'_{2} = \frac{W_{2}}{W} = \frac{1}{t}$.

So $u_{1} = -\int \frac{1}{t} dt = -ln|t|$, and $u_{2} = \int \frac{1}{t} dt = ln|t|$.

Hence, $y_{p} = -ln|t| e^{-t} + ln|t|te^{-t}$.

So our general solution, $$y = y_{c} + y_{p} = e^{-t}(c_{1}+c_{2}t-ln|t|+ln|t|t).$$

Does it look okay?

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  • $\begingroup$ Generally you only do particular solutions if it's a polynomial times a sine/cosine times an exponential, not a rational function. I'd use variation of paramaters $\endgroup$ – Alan Feb 5 '15 at 1:55
  • $\begingroup$ @Alan I haven't learned that technique yet. Is it the method of undetermined coeffecients? $\endgroup$ – OGC Feb 5 '15 at 1:57
  • $\begingroup$ What techniques have you learnt? $\endgroup$ – mattos Feb 5 '15 at 2:22
  • $\begingroup$ @Mattos Separation of variables technique and learned method of undetermined coefficients today. Seeing youtube videos on change of var technique. Is it possible to solve this problem by the method of undetermined coefficients technique? $\endgroup$ – OGC Feb 5 '15 at 2:26
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    $\begingroup$ @user36829 If you want, I'll post a solution using variation of parameters. $\endgroup$ – mattos Feb 5 '15 at 3:50
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You have already got

$$y_h = c_1e^{-t} + c_2te^{-t}$$

with

$$y_1 = e^{-t} \ \ \ \ \ y_2 = te^{-t}$$

Taking the Wronskian, you also found

$$W = e^{-2t}$$

The particular solution is given by the formula

$$y_p = -y_1 \int \frac{y_2 g(t)}{W} dt + y_2 \int \frac{y_1 g(t)}{W} dt$$

where

$$g(t) = \frac{e^{-t}}{t^{2}} = t^{-2}e^{-t}$$

Hence

$$\begin{align} y_p &= -e^{-t} \int \frac{te^{-t} \cdot t^{-2}e^{-t}}{e^{-2t}} dt + te^{-t} \int \frac{e^{-t} \cdot t^{-2}{e^{-t}}}{e^{-2t}} dt \\ &= -e^{-t} \int \frac{1}{t} dt + te^{-t} \int \frac{1}{t^{2}} dt \\ &= -e^{-t} \ln(t) - \frac{te^{-t}}{t} \\ &= -e^{-t}(\ln(t) + 1) \\ \end{align}$$

Therefore,

$$\begin{align} y &= y_h + y_p \\ &= e^{-t}(c_1 + c_2 t - \ln(t) - 1) \\ \end{align}$$

You can check by differentiation that this satisfies the ODE.

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  • $\begingroup$ $-1\cdot e^{-t}$ is an unnecessary part of the solution since it solves the homogeneous equation. I.e. the $-$ can be 'absorbed' into the $c_1$ coefficient. $\endgroup$ – jdods Mar 5 '15 at 1:58
  • $\begingroup$ @jdods Yeah and? $\endgroup$ – mattos Mar 5 '15 at 8:38
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In this case of only one characteristic value, also a change of function can help. Substitute, according to the homogeneous solution, $y(t)=e^{-t}u(t)$ to get for $u$ $$ u''(t)=t^{-2}\implies u'(t)=-t^{-1}+C\implies u(t)=-\ln|t|+Ct+D $$ and thus $$ y(t)=e^{-t}(-\ln|t|+Ct+D) $$

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I'll use the formula I developed here $$Y_p=y_h\int \left\{y_h^{-2} W_0 \left(\int y_h W_0^{-1} g \ dt\right)\right\} \ dt$$ where the Wronskian (up to a constant multiple) is $$W_{0}=\exp\left(-{\int p \ dt}\right)$$ and a solution to the homogeneous equation is $y_h=e^{-t}$ (we can use either).

So we first integrate $$W_{0}=\exp\left(-{\int 2 \ dt}\right)=e^{-2t}$$ and then we'll find the particular solution by $$ \begin{aligned} Y_p&=e^{-t}\int \left\{e^{2t} e^{-2t} \left(\int e^{-t} e^{2t} \frac{e^{-t}}{t^2} \ dt\right)\right\} \ dt \\ &=e^{-t}\int \left(\int \frac{1}{t^2} \ dt\right) \ dt \\ &=e^{-t}\int \left(-\frac{1}{t}\right) \ dt \\ &=-e^{-t}\ln t. \end{aligned} $$

This is a nice formula because you only need one solution to the homogeneous equation.

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