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I need to solve this equation. Please help me!

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

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

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2 Answers 2

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.

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Hint: What functions do you know whose second derivative is proportional to the function? There should be a two-dimensional solution space to the homogeneous equation (where the RHS is set to zero)

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