Multiple Scale Expansion: Slowly Varying Coefficients. In slowly varying coefficients our differential equation is: 

$$y''(t)+k^2(\epsilon t)y(t)=0~\text{ where }~ y(0)=a, y'(0)=b.$$ 

The textbook "Introduction to Perturbation Methods" says that we still have to determine the fast oscillator, and so we consider $t_1=f(t,\epsilon)$ and $t_2=\epsilon t$. When we apply chain rule to these we arrive at 
$$
\frac{d}{dt}=\frac{\partial f}{\partial t}\frac{\partial}{\partial t_1}+\epsilon\frac{\partial f}{\partial t_2}  ;
\\
\frac{d^2}{dt^2}=f_t^2\frac{\partial}{\partial t_1}+f_{tt}\frac{\partial}{\partial t_1}+2\epsilon f_t\frac{\partial ^2}{\partial t_1 \partial t_2}+\epsilon^2 \frac{\partial ^2}{\partial t_2^2}.$$
After substituting this into the main equation, we get
$$(f_t^2\partial_{t_1}^2+f_{tt}\partial_{t_1}+2\epsilon \partial^2_{t_1t_2}+\epsilon^2\partial^2_{t_2})y+k^2(\epsilon t)y=0$$
My question lies in the above equation. How do we say that the first and the last term balance each other, i.e., the $f_t^2\partial_{t_1}^2$ and $k^2(\epsilon t)$?
thank you!
 A: I'm not sure if I would call this "balancing", but in introducing an auxiliary function $f$ you got a degree of freedom in the collection of functions $k,y,f$ that can be fixed to any functional relation. Additionally you gain a degree of freedom by splitting the time scales. So selecting to set the sum of these two terms to zero fixes, removes one degree of freedom and leaves you with two new equations
$$
f^2_t∂^2_{t_1}Y(t_1,t_2)+k(t_2)^2Y(t_1,t_2)=0\\
f_{tt}\partial_{t_1}Y(t_1,t_2)+2\epsilon f_t \partial^2_{t_1t_2}Y(t_1,t_2)+\epsilon^2\partial^2_{t_2}Y(t_1,t_2)=0
$$
The other degree of freedom is used up by setting $f_t(t,ϵ)=k(ϵt)$, so that with $K'=k$ you get $t_1=f(t,ϵ)=K(ϵt)/ϵ$ and thus
$$
Y(t_1,t_2)=c_1(t_2)\cos(t_1)+c_2(t_2)\sin(t_1)
$$
Then the second equations gives in the coefficients of $\cos(t_1)$ and $\sin(t_1)$ the conditions
$$
k'(t_2)c_2(t_2)+2k(t_2)c_2'(t_2)+ϵc_1''(t_2)=0
\\
k'(t_2)c_1(t_2)+2k(t_2)c_1'(t_2)-ϵc_2''(t_2)=0
$$
so that in first order $c_1(t_2)=\frac{d_1}{\sqrt{k(t_2)}}$ and $c_2(t_2)=\frac{d_2}{\sqrt{k(t_2)}}$.
