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 $$y(t)=y_c(t)+y_p(t) $$


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


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.

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  • $\begingroup$ Please let me know how I can improve my answer. I really want to give you the best answer I can. $\endgroup$ – Mark Viola May 9 '18 at 17:53
  • $\begingroup$ It's so sweet. But I haven't learned Green function. It's on my textbook but not in the learning plan. I need one or two weeks to absorb this. $\endgroup$ – Jaqen Chou May 10 '18 at 3:43
  • $\begingroup$ OK. Well let me know if you have any questions. And feel free to up vote and accept an answer as you see fit. ;-) $\endgroup$ – Mark Viola May 10 '18 at 13:40

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.

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  • 1
    $\begingroup$ One can obtain an "exact" solution in terms of an integral expression in which the kernel is the Green (or Green's) Function. $\endgroup$ – Mark Viola May 9 '18 at 17:14
  • $\begingroup$ yeah, I also thought about just write it as a form of integral. But I am not so sure. Thank you for the confirmation. $\endgroup$ – Jaqen Chou May 9 '18 at 17:21

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