0
$\begingroup$

While trying to solve this differential equations I got stuck, at the very begging where I have these bunches of indefinite integrals whom I don't know how to solve. A friend of mine told me that is very easy to solve, because it's just basic exponential integrals. Well it's easy for him to say, because he is studying maths unlike me. The differential equation is: $$y''-6y'+9y = \frac{9(x+1)^2+6(x+1)+2}{(x+1)^3}$$ I tried to solve this by using the Wronskian method, even if try any other method I get the same indefinite integrals, where I get: $$y(x)= C_1e^{3x}+C_2(xe^{3x})$$ For $e^{3x}$ I get $$-e^{3x} \int x\frac{9(x+1)^2+6(x+1)+2}{(x+1)^3e^{3x}}dx$$ By separating them, I get 3 indefinite different integrals that I don't know hot to solve.... \begin{align} \ & -9e^{3x} \int \frac{x}{(x+1)e^{3x}}dx \\ \ &-6e^{3x} \int \frac{x}{(x+1)^2e^{3x}}dx \\ \ &-2e^{3x} \int \frac{x}{(x+1)^3e^{3x}}dx \\ \end{align} I guess that I should use integration by parts...

$\endgroup$

closed as unclear what you're asking by JonMark Perry, Michael Burr, José Carlos Santos, Shailesh, Namaste Jan 21 '18 at 22:05

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

1
$\begingroup$

$$y''-6y'+9y = \frac{9(x+1)^2+6(x+1)+2}{(x+1)^3}$$ Now let's consider $g(x)=\frac 1 {(x+1)}$

Then: $$g'(x)=\frac {-1}{(x+1)^2}, g''(x)=\frac 2 {(x+1)^3}$$ So, we have: $$y''-6y'+9y =9g(x)-6g'(x)+g''(x)$$ $$\underbrace{(y''-g'')}_\color{red}{z''}-6\underbrace{(y'-g')}_\color{red}{z'}+9\underbrace{(y-g) }_\color{red}{z}=0$$ Substitute $z(x)=y(x)-g(x)$ to get the homogenous equation: $$z''-6z'+9z = 0$$

Now it's easy to solve without any integration... $$r^2-6r+9=0 \implies r=3$$ $$z(x)=K_1e^{3x}+K_2xe^{3x}$$ $$\boxed{y(x)=K_1e^{3x}+K_2xe^{3x}+\frac 1 {(x+1)}}$$

$\endgroup$
  • 1
    $\begingroup$ Thank you! Could you just explain me how did you use the substitution z(x)=y(x)-g(x) to get z''-6z'+9z=0 equation ? $\endgroup$ – oootelecs Jan 21 '18 at 0:56
  • $\begingroup$ I edited my answer I hope it is more clear now.... for the substitution.. $\endgroup$ – Isham Jan 21 '18 at 1:38
  • 1
    $\begingroup$ Yes it is! Thank you!!! $\endgroup$ – oootelecs Jan 21 '18 at 7:12

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