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# 14.9: A Second Order Differential Equation

Solve

$\ddot y - 4 \dot y + 3y = e^{-t} \label{eq1}$

with initial conditions $$y_0=1$$ and $$\dot y_0 = -1$$.

You probably already know some method for solving this equation, so please go ahead and do it. Then, when you have finished, look at the solution by Laplace transforms.

Laplace transform:

$s^2 \bar{y} - s + 1 - 4(s \bar{y} - 1) + 3\bar{y} = 1/(s+1).$

(My! Wasn't that fast!)

A little algebra:

$\bar{y} = \frac{1}{(s-3)(s-1)(s+1)}+\frac{s-5}{(s-3)(s-1)}.$

Partial fractions:

$\bar{y}=\frac{1}{8}\left(\frac{1}{s-3}-\frac{2}{s-1}+\frac{1}{s+1}\right) + \frac{2}{s-1}-\frac{1}{s-3},$

or

$\bar{y}=\frac{1}{8} \left(\frac{1}{s+1}\right) + \frac{7}{4}\left(\frac{1}{s-1}\right) - \frac{7}{8}\left(\frac{1}{s-3}\right).$

Inverse transforms:

$y=\frac{1}{8}e^{-t} + \frac{7}{4}e^t - \frac{7}{8}e^{3t}$

and you can verify that this is correct by substitution in the original differential equation (Equation \ref{eq1}).

So: We have found a new way of solving differential equations. If (but only if) we have a lot of practice in manipulating Laplace transforms, and have used the various manipulations to prepare a slightly larger table of transforms from the basic table given above, and we can go from $$t$$ to $$s$$ and from $$s$$ to $$t$$ with equal facility, we can believe that our new method can be both fast and easy.

But, what has this to do with electrical circuits? Read on.