Find the power series solution about $x=0$ of the differential equation $y''-y=x$.

Question

My attempt

I know how to solve $y''-y=0$ and after solving I get$$ y = a_o(1+ x²/2 + x⁴/24 + x^6/720 + ....) + a_1(x + x³/6 + x^5/120 + ....).$$ But I don't know how to solve $y''-y=x.$

In my book answer is $$y = a_o(1+ x²/2 + x⁴/24 + x^6/720 + ....) + a_1(x + x³/6 + x^5/120 + ....) + (x³/6 + x^5/120 + ....).$$

Please give me some hint

Answer 1

Consider the function$$y(x)=\frac{x^3}{3!}+\frac{x^5}{5!}+\frac{x^7}{7!}+\cdots$$Then$$y''(x)=x+\frac{x^3}{3!}+\frac{x^5}{5!}+\cdots$$and therefore $y''-y=x$. If $z$ is another solution of the equation $y''-y=x$, then$$(z-y)''-(z-y)=z''-z-(y''-y)=x-x=0,$$and therefore $z-y$ is of the form$$a_0\left(x+\frac{x^3}{3!}+\frac{x^5}{5!}+\cdots\right)+a_1\left(1+\frac{x^2}{2!}+\frac{x^4}{4!}+\cdots\right).$$So,$$z(x)=a_0\left(x+\frac{x^3}{3!}+\frac{x^5}{5!}+\cdots\right)+a_1\left(1+\frac{x^2}{2!}+\frac{x^4}{4!}+\cdots\right)+\frac{x^3}{3!}+\frac{x^5}{5!}+\frac{x^7}{7!}+\cdots$$

Answer 2

You can rewrite this specific equation as $$ (y+x)''-(y+x)=0 $$ and already know the solution for $y+x$. That your form looks more complicated is due to using $a_1=y'(0)$ as parameter.