# Solution for the equation $\frac{-\mathrm d^2u}{\mathrm dt^2}+\frac{u}{LC}=\frac{V}{LC}$

$$\dfrac{-\mathrm d^2u}{\mathrm dt^2}+\dfrac{u}{LC}=\dfrac{V}{LC},\qquad L,C,V,I=\text{constant}$$

and $u(0)=0$, $\left. \dfrac{\mathrm du}{\mathrm dt}\right\vert_{t=0}=\dfrac{I}{C}$

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Here are some hints. It's a standard form of 2nd order ordinary differential equation. You first find a "particular solution" $u_p$ to your equation (not including your conditions on $u(0)$ and ${du \over dt}(0)$). In this case it's just a constant solution. Then you find the general solution to the homogeneous equation $-{d^2 u \over dt} + {1 \over LC} u = 0$. This will be of the form $a u_1(t) + b u_2(t)$ for the right two functions, which will be certain exponentials here.
Then any solution to $-{d^2 u \over dt} + {1 \over LC} u = {V \over LC}$ will be of the form $a u_1(t) + b u_2(t) + u_p(t)$. Plug in the conditions on $u(0)$ and ${du \over dt}(0)$ to figure which $a$ and $b$ correspond to your situation. Voila, you're done.