Find a particular solution for $y''+2y'+2y=\ln(t)$

Question

It's easy to know that the ode's complementary solutions are $$e^{-t}\cos t, e^{-t}\sin t$$ hence we can write the general solution as $$y(t)=c_1(t)\cdot e^{-t}\cos t+c_2(t)\cdot e^{-t}\sin t $$ And there comes $$c_1'(t)\cdot e^{-t}\cos t+c_2'(t)\cdot e^{-t}\sin t=0 \\ c_1'(t)\cdot e^{-t}(-\sin t- \cos t)+c_2'(t)\cdot e^{-t}(\cos t-\sin t)=\ln x$$ By solve $c_1'(t)$ and $c_2'(t)$, I get $$c_1'(t)=-e^{t}\ln t\cdot\sin t,c_2'(t)=e^{t}\ln t\cdot\cos t $$ But I can't integrate it to get an exact solution.
So is there any other way to get a particular solution such that I can write the solution as $$y(t)=y_c(t)+y_p(t) $$

Answer 1

The antiderivative $\int e^t\sin t\ln tdt$ or its $\cos$ counterpart is known to be not expressible as a combinations of elementary functions you have in mind. But you have basically solved your problem - there is no problem in writing $\int_{t_0}^tf(s)ds$ for the antiderivative of any integrable function $f(t)$.

With that said, this seems like a problem from an elementary ODE class, and in such classes the functions on the right hand side are usually a combination of the exponentials, trig functions and polynomials. In that case, you are guaranteed to be able to integrate any function that come your way using the method of variation of parameters.

Answer 2

We can develop a Green (or Green's) Function for the ODE

$$G''+2G'+2G=0$$

for $t\ne t'$ with boundary conditions $G(0,t')=G(T,t')=0$, $(0<T<\pi)$ and

$$\left.\frac{\partial G(t,t')}{\partial t}\right|_{(t=t'^+)}-\left.\frac{\partial G(t,t')}{\partial t}\right|_{(t=t'^-)}=1 \tag1$$

Then, we have

$$G(t,t')=\begin{cases}-\frac{\sin(T-t')\sin(t)}{\sin(T)}\,e^{-(t-t')}&,0\le t<t'\\\\ -\frac{\sin(t')\sin(T-t)}{\sin(T)}\,e^{-(t-t')}&,t'<t\le T\tag2 \end{cases}$$

Equipped with $(2)$, we find that

$$y_p(t)=\int_0^T G(t,t')\log(t')\,dt'$$

where $y_p''(t)+2y_p(t)+2y(y)=\log(t)$, $y_p(0)=y_p(T)=0$.

And we are done.