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!
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 $$\int(1+2\ln(t))\,t^{m}\,dt=\frac{t^{m+1} }{(m+1)^2}(2 (m+1) \log (t)+m-1)$$