Question

d²V/dt² -4V = 8e^(2t)

When t=0 and V=4 and dV/dt = 6

Answer 1

This is a second order differential equation with constant coefficients. The characteristic polynomial of the homogeneous equation is $r^2-4=0$. The roots of this polynomials are $r_1=2 \text{ and } r_2=-2$. The general solution of the homogeneous equation is then $V = c_1e^{2t}+c_2e^{-2t}$.

A particular solution of the nonhomogeneous equation is $cte^{2t}$, where $c$ is a constant. To determine it, we replace $V=cte^{2t} \text{ and }\frac{d^2V}{dt^2}=4c(1+t)e^{2t}$ in the equation, we obtain

$4c(1+t)e^{2t}-4cte^{2t}=8t\implies 4ce^{2t}=8e^{2t}\implies c=2.$

The solution is then

$V(t) = c_1e^{2t}+c_2e^{-2t}+2te^{2t}.$

To determine $c_1\text{ and }c_2$ we use the initial conditions:

$V(0) = c_1+c_2=4$ and

$\frac{dV}{dt}=(2c_1+4t+2)e^{2t}-2c_2e^{-2t}\Rightarrow\frac{dV}{dt}(0)=2c_1-2c_2+2=6\Rightarrow c_1-c_2=2$.

Then $c_1=3\text{ and }c_2=1$. We obtain finally

$V(t)=3e^{2t}+e^{-2t}+2te^{2t}.$