# How to solve these indefinite integrals (trying to solve differential equation)? [closed]

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...

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

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$$y''-6y'+9y = \frac{9(x+1)^2+6(x+1)+2}{(x+1)^3}$$ 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... $$r^2-6r+9=0 \implies r=3$$ $$z(x)=K_1e^{3x}+K_2xe^{3x}$$ $$\boxed{y(x)=K_1e^{3x}+K_2xe^{3x}+\frac 1 {(x+1)}}$$