Not a homework question, just a high schoolers interested in quantum mechanics.
Given the differential equation: $$\frac{d^2 \psi }{d x^2}=\frac{-2mE}{\hbar^{2}}\psi$$
How would you go on solving this differential through double integration?
Assume $\psi$ is with respect to $x$...
Now, of course, the equation can be solved by guessing the $\psi$. Since $\frac{d^2 \psi }{d x^2}, $ is proportional to $\psi$, the $\psi$ can be guessed to be exponential as in $\psi=Ae^{kx}$ and work it out accordingly. But I was thinking if there was a way to solve the differentiation equation above through integrating the differential twice and do it.
Again, it's not a homework question. I was self-studying quantum myself and I came to this equation and got stuck. Anyways, any help will be appreciated!
According to your interest " if there was a way to solve the differentiation equation above through integrating the differential twice and do it", you can proceed as follows
Here the differential equation is$$\frac{d^2 \psi }{d x^2}=\frac{-2mE}{\hbar^{2}}\psi\tag1$$ Clearly here the term $~\frac{2mE}{\hbar^{2}}~$ is independent of $~x~$, so we take equation $(1)$ as $$\frac{d^2 \psi }{d x^2}=-~k~\psi$$where $~k=\frac{2mE}{\hbar^{2}}=\text{constant}~.$
Multiplying both side of the above equation by $~2\frac{d \psi }{d x}~$, we have $$2\frac{d \psi }{d x}\cdot\frac{d^2 \psi }{d x^2}=-~2~k~\psi\cdot\frac{d \psi }{d x}~$$ $$\implies \frac{d }{d x}\left[\left(\frac{d \psi }{d x}\right)^2\right]=-~\frac{d }{d x}~\left(k~\psi^2\right)$$ Integrating we have,$$\left(\frac{d \psi }{d x}\right)^2=-~k~\psi^2~+~c$$where $~c~$ is a constant. $$\implies \frac{d \psi }{d x}=\pm ~\sqrt{c~-~k~\psi^2}$$ Again integrating we have,$$\int\dfrac{d\psi}{\sqrt{c~-~k~\psi^2}}~=~\pm~\int dx~+~d\tag2$$Now if $~k\gt 0~$, the equation $(2)$ gives $$ \dfrac{1}{\sqrt k}~\sin^{-1}\left(\dfrac{\sqrt k}{\sqrt c}~\psi\right)~=~\pm~x~+~d$$ $$\implies \psi~=~\dfrac{\sqrt c}{\sqrt k}~\sin\left[\sqrt k~(\pm~x~+~d)\right]\tag3$$where $~c~$ and $~d~$ are constants.
Also if $~k\lt 0~$, the equation $(2)$ gives $$ \dfrac{1}{\sqrt k}~\sinh^{-1}\left(\dfrac{\sqrt k}{\sqrt c}~\psi\right)~=~\pm~x~+~d$$ $$\implies \psi~=~\dfrac{\sqrt c}{\sqrt k}~\sinh\left[\sqrt k~(\pm~x~+~d)\right]\tag4$$where $~c~$ and $~d~$ are constants.
Note: The right hand side terms in equation $(3)$ and $(4)$ can be transformed (by using trigonometric rules) in such a way that after a few steps we can get the following
For $~k\gt 0~$, equation $(3)$ becomes $$\psi=A_1\sin\left(\sqrt k~x\right)~+~B_1\cos\left(\sqrt k~x\right)$$ For $~k\lt 0~$, equation $(4)$ becomes $$\psi=A_2\sinh\left(\sqrt k~x\right)~+~B_2\cosh\left(\sqrt k~x\right)$$ where $~A_1,~B_1,~A_2,~B_2~$ are constants and $~k=\frac{2mE}{\hbar^{2}}~$.