Solving a generic second order differential equation Ok guys, I have to solve this ODE
$$
\frac{d^2y}{dx^2}=f(x), \quad 
x>0,\quad y\left(0\right) = 0, \quad 
\left.\frac{dy}{dx}\right\lvert_{x=0}=0 
$$
The solution I should get is in the form of 
$$y\left(x\right)=\int_0^x k\left(t\right)\, dt  $$
Moreover, I should tell what the function $\,k\left(t\right)\,$ is in a simple form. I have tried by substitution, with $\,u=y',\,$ but I have to integrate two times $\,f\left(x\right)\,$, and this seems not at all like a simple form... The function $\,k\left(t\right)\,$ is undoubtedly an exponential, but in which form? Hope that somebody can help!
 A: The Green's function for $\dfrac{d^2}{dx^2}$ is the ramp function $x\theta(x)$, where $\theta(x)$ is the Heaviside step function. We can write the solution for the general inhomogeneous problem as follows. $$\begin{align*}y(x)&=\int_{-\infty}^\infty (x-\chi)\theta(x-\chi)f(\chi)d\chi\\&=\int_{-\infty}^x (x-\chi) f(\chi)\, d\chi\end{align*}$$for suitable conditions on $f$.
A: You are given a form that will do the trick. The thing you need to do now is to differentiate it twice. By comparing result with your equation you will get condition on function $k$ that will make it work. This might involve double integration but you can try to simplify it by changing the order of integrals. This way you can make it a single integral (over slightly modified function)
A: HINT: Use the (Leibniz) formula for differentiation under the integral with variable limits
$$
\frac{d}{dx}\left( 
\int_{a\left(x\right)}^{b\left(x\right)} f\left(x,t\right)\, dt  \right)
= f\big(x,b\left(x\right)\big) \cdot b'\left(x\right)
- f\big(x,b\left(x\right)\big) \cdot a'\left(x\right)
+ \int_{a\left(x\right)}^{b\left(x\right)} 
\frac{\partial}{\partial x}\,f\left(x,t\right)\, dt 
$$

In your particular case we have
$$
y\left(x\right) = \int_0^x k\left(t\right)\, dt 
\implies
\frac{dy}{dx} = k\left( x\right) \cdot 1 - 0\cdot 0  +  \int_0^x 0\, dt  = k\left( x\right)
\implies
\frac{d^2y}{dx^2} = k'\left( x\right)
$$
Substituting this back into original equation we get
$$
k'\left(x\right)=f\left(x\right), \qquad k\left(0\right)=0 
$$
After expressing $\,k\left( x\right)\,$ via $\,f\left( x\right)\,$ you can find solution for $\,y\,$ in terms of $\,f.\,$
Hope you can proceed from here.
