# How are the initial conditions transformed with a change of variable?

I have an ODE of the form $y''(x)=F(x,y')$ which has the initial conditions, $y(\mu)=\mu$ and $y'(\mu)=1$. Now I have seen that an equation which is equidimensional in x, can be made autonomous by the change of variable $x=e^t$, ie now we must solve an ODE for the unknown $z(t)$ ($=y(x)$). But now what happens to the initial conditions??

I often see change of variables of some sort in books. But no one really talks about what happens to the initial conditions.

I guess people dont need to transform the initial conditions, since if a geneneral solution $z(t)$ to the transformed (and simplified) equation can be found. Then we can resolve any arbitrary constants, using the fact $z(ln(t))=y(x)$. But I'm trying to solve my equation through power series techniques so I dont think this method applies. I need the conditions in advance to compute the coefficients of the terms in the power series.

I think the chain rule may be applied to evaluate one of the conditions. That is by application of the chain rule we have $\frac{dy}{dx}=\frac{dy}{dt}\frac{1}{t}$, so $\frac{dy}{dt}(ln(\mu))=\mu$ since $\frac{dy}{dx}(\mu)=1$. But what about the first condition? how is that transformed?

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It is much like changing variables in definite integrals. You can either change the old conditions to the new variables, or change the solution back to the old variables and use the original conditions. Say your initial conditions are $y(1)=\mu, y'(1)=1$ (you shouldn't have a variable as the argument of $y$ for initial conditions-it is given). Then if you set $x=e^t$ you have $y(t=0)=\mu, \frac{dy}{dt}(t=0)=\frac{dy}{dx}\frac{dx}{dt}(t=0)=1\cdot 1=1$ if you consider $y$ as a function of $t$