Consider equations defined in $D_x = (0,\infty) \times (0,\infty)$
1) Find the maximal solution to
$$ \dot y = \frac{3t^2}{5y^4} $$
satisfying the initial value problem $y(1) = 2^{\frac{1}{5}}$
I have found the maximum solution to be $((0,\infty), (t^3+1)^\frac{1}{5})$
2) Find the maximal solution to
$$ \dot y = -\frac{2y}{5t}+\frac{3}{5y^4}$$
satisfying the IVP $y(1) = 2^{\frac{1}{5}}$.
(HINT: Use a transformation of the form $t^{\alpha}y(t)$.
I hope someone can explain how to use the transformation to find the maximal solution.
Best Answer
Multiply both sides of $$\dot y = -\frac{2y}{5t}+\frac{3}{5y^4}$$ by $5y^4$ to get $$ 5y^4\dot y = -\frac {2}{t} y^5 +3$$
Now let $$u=y^5$$ and your equation transforms to $$ \dot u = -\frac {2}{t} u +3 $$
which is a linear equation.
Solve for $u$ and back substitute to find $y$