Solving a RLC-series circuit with Wolfram Alpha?

Questions
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1

For solving a RLC-series circuit with R= 5 ohms, L = 0,03 H , C = 1 microF , null initial conditions and voltage = 10 V, I tried with this syntax on the Wolfram alpha , but it does not work…
10=5x(t) + 0.03dx(t)/dt + (10^1)*Integrate[x(t),t] ; x(t=0)=0; x´(t=0)=10/0.03
What may be wrong, folks ?

2

I myself don’t understand it. The integral has no bounds. You use a mixture of different notations for the derivative.

3

Thank you for your reply, Gianluca. In fact, the value of the capacitor is 0,1 F, in accordance with my equation. In this case, I consider x(t) as being the instantaneous value of the current. Regarding the integral limits, I confess I don´t know how to set them.