To solve the differential equation use an appropriate change of variables.

Question

$$2t\left(e^2\right)^y \frac{dy}{dt}=3t^4+\left(e^2\right)^y$$

Please help. I'm not sure where to start.

Answer 1

HINT

The obvious one is to first try $u = e^{2y} = \left(e^2\right)^y$ and then $u' = e^{2y}\cdot 2y' = 2uy'$...

Answer 2

HINT

Substitute $u=e^{2y}$ - and therefore $u'=2y'e^{2y}$ - to get an ODE of the form

$$tu'=3t^4+u\Leftrightarrow u'-\frac1t u = 3t^3$$

which can be solved by standard techniques. To be exact first of all solve the homogenous DE $u'-\frac1t u = 0$ which leads to $u= c t$. Variation of constants yields to $c't=3t^3$ therefore $c(t)=t^3+k$ and so finally the solution for $u$ is given by $u=kt+t^4$. Resubstitution leads to $y(t)=\frac12\ln(kt+t^4)$. Everything clear now?

Answer 3

$$2t\left(e^2\right)^y \frac{dy}{dt}=3t^4+\left(e^2\right)^y$$ $$t(2e^{2y}y')=3t^4+\left(e^2\right)^y$$ $$t(e^{2y})'-e^{2y}=3t^4$$ Divide by $t^2$ both side $$\frac {t(e^{2y})'-e^{2y}}{t^2}=3t^2$$ Remember that $(\frac fg)'=\frac {f'g-fg'}{g^2}$ $$(\frac {e^{2y}}{t})'=3t^2$$ Now just integrate both side $$\frac {e^{2y}}{t}=3\int t^2dt$$