1
$\begingroup$

Would solving $ \dfrac{ d^2x }{ dt^2 } + 6 \dfrac{ dx }{ dt } + 9x = 4t^2 + 5 $ using variation of parameters require integration by parts or can I solve it without knowing integration by parts?

I'm not sure if I'm just using the method wrong or if it requires integration by parts. I'm new to variation of parameters, and I haven't encountered integration by parts. Thanks.

The non homogeneous characteristic equation is $ r^2 + 6r + 9 = 4t^2 + 5 $

The characteristic homogeneous equation is $ r^2 + 6r + 9 = 0 $

$ \Rightarrow (r + 3)(r + 3) $

$ \Rightarrow r = -3 $

$ \therefore x(t) = C_1e^{-3t} + C_2te^{-3t}$ is the general solution for the homogeneous differential equation.

I now find the particular solution.

$-e^{-3t} \displaystyle\int \dfrac{te^{-3t}(4t^2 + 5)}{W(x_1, x_2)} dt + te^{-3t} \displaystyle\int \dfrac{e^{-3t}(4t^2 + 5)}{W(x_1, x_2)} dt$

The Wronskian $W(x_1, x_2) = e^{-6t} $

$-e^{-3t} \displaystyle\int \dfrac{te^{-3t}(4t^2 + 5)}{W(x_1, x_2)} dt + te^{-3t} \displaystyle\int \dfrac{e^{-3t}(4t^2 + 5)}{W(x_1, x_2)} dt $ $= -e^{-3t} \displaystyle\int \dfrac{te^{-3t}(4t^2 + 5)}{e^{-6t}} dt + te^{-3t} \displaystyle\int \dfrac{e^{-3t}(4t^2 + 5)}{e^{-6t}} dt$

$\endgroup$
  • 1
    $\begingroup$ Could you type up some of your work? $\endgroup$ – Jacky Chong Oct 26 '16 at 1:21
  • 1
    $\begingroup$ Yes. You do need integration by parts. $\endgroup$ – Jacky Chong Oct 26 '16 at 1:34
  • 1
    $\begingroup$ I think you want "$\mathrm{e}^{-6t}$" in those last two denominators. $\endgroup$ – Eric Towers Oct 26 '16 at 1:36
  • 1
    $\begingroup$ It should be $e^{-6t}$ on the last line. Other than that, everything seems okay. $\endgroup$ – Jacky Chong Oct 26 '16 at 1:36
  • 1
    $\begingroup$ Yes. Simplify your integrand then use integration by parts. $\endgroup$ – Jacky Chong Oct 26 '16 at 1:39
0
$\begingroup$

Considering the general case of $$I_k=\int t^k e^{r t}\,dt$$ Using integration by parts $$u=t^k \implies du=k t^{k-1}\,dt$$ $$dv=e^{r t}\,dt\implies v=\frac{e^{r t}}{r}$$ makes $$I_k=\frac{t^k e^{r t}}{r}-\frac k r\int t^{k-1} e^{r t}\,dt=\frac{t^k e^{r t}}{r}-\frac k r I_{k-1}$$ with $I_0=\frac{e^{r t}}{r}$.

If you already heard about the incomplete gamma function, almost from definition, you would have $$I_k=-\frac{t^{k+1}} {(-r t)^{k+1}} \Gamma (k+1,-r t)$$ which, in the case where $r<0$ reduces to $$I_k=-\frac{\Gamma (k+1,-r t)} {(-r )^{k+1}} $$

$\endgroup$
0
$\begingroup$

$$ \int e^{r t} \sum_{i=0}^na_it^i dt = \frac{1}{D} e^{rt} \sum_{i=0}^na_it^i = e^{rt} \frac{1}{D+r}\sum_{i=0}^na_it^i = e^{rt}\sum_{i=0}^n \frac{a_i}{r} \frac{1}{1+D/r}t^i = $$ $$ e^{rt}\sum_{i=0}^n \frac{a_i}{r} \sum_{j=0}^{\infty} \left(-\frac{D}{r}\right)^j t^i = e^{rt}\sum_{i=0}^n \frac{a_i}{r} \sum_{j=0}^{i} \left(-\frac{D}{r}\right)^j t^i = e^{rt}\sum_{i=0}^n \frac{a_i}{r} \sum_{j=0}^{i} \frac{(-1)^j}{r^j} \frac{i!}{(i-j)!} t^{i-j} $$

$\endgroup$

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.