The equation I have is
$$ t^2(y')+2t(y)-(y^3)=0 $$
I am told that I should use the substitution $ z =2/y^2 .$
How do I get the $z$ equivalents for (I am assuming) $y'$ and $y$ ? I managed to get the substitution $z' = -1/y^3$ , but have difficulty getting the other 2.
And upon getting the 2 values, am I correct to say I need to sub it in, and put it in the form
$$ z' + p(x)z = q(x) $$
where $p(x)$ and $q(x)$ are functions
Any help is appreciated thank you!
On
The equation is of Bernoulli type, the substition $z=y^{-3}$ does not give the right degrees, you need to use $z=y^{-2}$, adding a constant factor to that is more of a convenience than necessary.
As a control, you could also just combine the two first terms to get $$ (t^2y(t))'-y(t)^3=0, $$ which is a separable equation leading to $$ \frac{(t^2y)'}{(t^2y)^3}=\frac1{t^6}. $$
Given $z = 2/y^2$, when you take the derivative you should get $z'$ in terms of $y'$ and so forth. Using the chain rule, $$z' = -\frac{4}{y^3}\cdot y'$$ and since $$y = \sqrt{\frac{2}{z}}$$, you can substitute to get $$t^2(-y^3z'/4) + 2t \sqrt{\frac{2}{z}}- (\sqrt{\frac{2}{z}})^3=0$$ and so $$t^2(-(\sqrt{\frac{2}{z}})^3z'/4) + 2t \sqrt{\frac{2}{z}}- (\sqrt{\frac{2}{z}})^3=0$$ and dividing out one factor of $\sqrt{\frac{2}{z}}$ should complete the substitution.