Convert given IVP $$y''-\sin x\ y'+e^xy=x \tag{1}$$ $y(0)=0;y'(0)=1 $ to the integral form.
My attempt:- I substitute $u(x)=y''(x)$. The we get on integration.
$$y'(x)=y'(0)+\int_{0}^x u(t)dt$$ $$y(x)=y(0)+x+\int_0^x \int_{0}^u u(t)dt du$$
We know that $\int_0^x \int_{0}^u u(t)dt du=\int_0^x (x-t)u(t)dt$.
So, our (1) becomes $$u(x)-\sin(x)( 1+\int_{0}^x u(t)dt)+e^x(x+\int_0^x (x-t)u(t)dt)=x $$
But Answer given in the text is by the substitution $y''=x+\sin x y'+e^x y$. Final answer was $y(x)=1-x+x^3/6+\int_0^x[\sin t-\cos t(x-t)-e^t(x-t)]y(t)dt$ which entirely different from my answer. Can you explain where is my error?
You have
$y(x)=y(0)+x+\int_0^x \int_{0}^x u(t)dt dt .$
For one thing, the variables of integration must have to be distinct, so this should be $y(x) =y(0)+x+\int_0^x \int_{0}^s u(t)dt ds $.
Second, the double integral should be (I think) $y(x) =y(0)+x+\int_0^x \int_{0}^s u(t)dt ds $.
With this, you can reverse the order of integration.