# Help with question: solve: $t x'=(1+2\ln(t)) \ , \tan(t)$, where $x≥ 0, t>0$

I've tried and asked everyone possible and I've done the rest of the questions in the assignment but I simply can't figure this one out. This section is differential equations.

The question is simply this: solve: $$tx'=(1+2 \ln(t)) \, \tan(t),$$ where $x≥ 0, t>0$.

I turned $x'$ into $\frac{dx}{dy}$ and mutated it to end up being $$\int dx =\int \frac{(1+2\ln(t))\,\tan(t)}{t}\,dt$$ But that can't be solved. So I literally have no clue as to solve this. Any and all help will be greatly appreciated!

• @Moo Yes this is the question straight from the assignment. And what do you mean by you would be stuck with numerical solutions? Jul 4 '17 at 3:37
• @Moo I'm sorry that I'm not understanding this completely, but I googled what that meant and it's things like Euler's rule and such (?) Is that what you are meaning? If so, what do you suggest I do. This is the first problem of this kind I've encountered and the rest has been fairly straight forward. Jul 4 '17 at 3:56
• @Moo Yeah that's what I think. Maybe there is a typo. However the last chapter we did learn "First order linear differential equations" that are of the form dy/dx + P(x)y = Q(x) but our prof said we would be tested on that next week and not this week but regardless I don't think this is that kind of problem either right? I couldn't get it in the form above anyway Jul 4 '17 at 4:38

Beside numerical integration, may be they would like to use the expansion $$\tan(t) = \sum^{\infty}_{n=1} \frac{B_{2n} (-4)^n \left(1-4^n\right)}{(2n)!} t^{2n-1}\qquad \text{for} \qquad |t| \lt \frac \pi 2$$ leading to $$\int\frac{(1+2\ln(t))\,\tan(t)}{t}\,dt=\sum^{\infty}_{n=1} \frac{B_{2n} (-4)^n \left(1-4^n\right)}{(2n)!}\int(1+2\ln(t))\,t^{2n-2}\,dt$$ and use integration by parts to get to $$\int(1+2\ln(t))\,t^{m}\,dt=\frac{t^{m+1} }{(m+1)^2}(2 (m+1) \log (t)+m-1)$$