For this question, I'm having trouble solving this equation. I'm not sure how to proceed further with the substitution. Here is what I have so far. Can anyone please help me out?
For this question, I'm having trouble solving this initial-value problem. Here is what I have so far. Can anyone please help me out?
$$ \frac{du}{dt} = \frac{{u}^{3/2} + \arctan(t)+2u}{1+t^2}, u(0)=4\\ \frac{du}{dt} = u^{3/2} (\frac{\arctan(t) + 2u^{-1/2}}{1+t^2}) $$ Let $x = u^{-1/2}\,dx = \frac{-1}{2}u^{-3/2} \,du$
The differential equation being $$u'=\frac{u^{3/2}+2 u+\tan ^{-1}(t)}{1+t^2}$$ consider first the homogeneous equation $$u'=\frac{u^{3/2}+2 u}{1+t^2}$$ which is separable and leads to $$\log \left(\frac{\sqrt{u}}{2+\sqrt{u}}\right)=\tan ^{-1}(t)+C$$ There are potentially two (nasty) solutions for $u$.
Now, I suppose that you need to define the integration factor.