Finding leading order approximation to $\frac{d^2y}{dx^2}-\epsilon y=x$

Question

The conditions provided were $y(0)=1$ and $y'(0)=1$.

Since this equation is regular, I can neglect the term involving $\epsilon$, giving $\frac{d^2y}{dx^2}=x$.

I tried to solve this in order to obtain the leading order approximation.

The characteristic equation is $m^2=0$, so the roots are $0$.

General solution: $y=A+Bx$

$y(0)=1$ implies $1=A$ and $y'(0)=1$ implies $1=B$. So $y=1+x$.

According to the solution, the leading order approximation is $\frac{1}{6}x^3+x+1$. If my thinking is correct, where did the extra $\frac{1}{6}x^3$ term come from? Or have I been mistaken somewhere?

Answer 1

You may solve $$\frac{d^2y}{dx^2}=x \implies \frac{dy}{dx}= \frac {1}{2}x^2+C_1$$ Which in term results in $$ y=(1/6)x^3+C_1x+C_2 $$

with the given initial condition we get $C_1=C_2=1$

Resulting in $$y= (1/6)x^3+x+1 $$

Answer 2

The solution of the DE is $a\sin (\sqrt {\epsilon})+b \cos (\sqrt {\epsilon})$ and the initial conditions give $a=-\frac 1 {\sqrt {\epsilon}}, b=1$. Expand $\sin $ and $\cos$ up to the term in $x^{2}$. You will get $1+x+\frac 1 6 x^{3}$.