IVP variable change. I'm working in this IVP:
$$(4t^2+16t+17)y'' - 8y = 0, \quad y(-2) = 1, \quad y'(-2) = 0$$
and it's asked to do this variable change $x=t+2$.
I did as asked and found the equation:
$$ (4x^2 + 1)y'' - 8y = 0. $$
Changing the $t \to x$: then $ x = t + 2 $.
Changing $y \to s$: then $ s = y - 1 $.
With these we can change this: $ y(-2) = 1 $ to $ s(0) = 0 $
but I'm having some trouble to change the $ y'(-2) = 0 $ to $ s $ end the question.
Any help is appreciated!
 A: From $$ (4 \, t^2 + 16 \, t + 17) \, y^{''}(t) - 8 \, y^{'}(t) = 0, \quad y(-2) = 1 \quad y^{'}(-2) = 0 $$ and knowing that the transformation $t \to x-2$ needs to be performed then:
$$ dt = dx \to \frac{dx}{dt} = 1$$ and
\begin{align}
\frac{dy}{dt} &= \frac{dx}{dt} \, \frac{dy}{dx} = \frac{dy}{dx} \\
\frac{d^2 y}{d t^2} &= \frac{d^2 y}{d x^2}.
\end{align}
Since $x = t + 2$ then when $t = -2$ this gives $x = 0$ such that $y(-2)=1$ becomes $y(0)=1$ when $t \to x$ and $y^{'}(-2) = 0$ becomes $y^{'}(0) = 0$. Now the differential equation becomes, in $x$ variable,
$$ (4 \, x^2 + 1) \, y^{''}(x) - 8 \, y(x) = 0, \quad y(0) = 1, \quad y^{'}(0) = 0.$$
This has the solution
$$ y(x) = c_{1} (4 \, x^2 + 1) + c_{2} \, ((4 \, x^2 + 1) \, \tan^{-1}(2 x) + 2 x). $$
Using $y(0) = 1$ gives $c_{1} = 1$. Using $y^{'}(0) = 0$ gives $c_{2} = 0$ and $$ y(x) = 4 \, x^2 + 1.$$ Returning to $t$ variable this gives the solution $$ y(t) = 4 \, (t + 2)^2 + 1.$$
Note: The original equation can be solved as is and gives the general solution
$$ y(x) = c_{1} \, (4 \, t^2 + 16 \, t + 17) + c_{2} ((4 \, t^2 + 16 \, t + 17) \, \tan^{-1}(2 t + 2) + 2 t + 4). $$
Applying the conditions $y(-2) = 1$ and $y^{'}(-2) = 0$ gives
$$ y(t) = 4 \, (t+2)^2 + 1 $$ as the solution.
