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# 14.8: A First Order Differential Equation

Example $$\PageIndex{1}$$

Solve $$\dot y + 2y = 3te^t$$ with initial condition $$y_0=0$$.

Solution

If you are in good practice with solving this type of equation, you will probably multiply it through by $$e^{2t}$$, so that it becomes

$\frac{d}{dt}\left(ye^{2t}\right) = 3te^{3t},$

from which

$$y = (t-\frac{1}{3})e^t + Ce^{-2t}.$$

(You can now substitute this back into the original differential equation, to verify that it is indeed the correct solution.)

With the given initial condition, it is quickly found that $$C=\frac{1}{3}$$ so that the solution is

$y = te^t -\frac{1}{3}e^t + \frac{1}{3} e^{-2t}.$

Now, here's the same solution, using Laplace transforms.

We take the Laplace transform of both sides of the original differential equation:

$s\bar{y} +2\bar{y} = 3 \textbf{L}(te^t) = \frac{3}{(s-1)^2}.$

Thus

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

Partial fractions:

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

Inverse transforms:

$y=\frac{1}{3} e^{-2t} - \frac{1}{3} e^t +te^t.$

You will probably admit that you can follow this, but will say that you can do this at speed only after a great deal of practice with many similar equations. But this is equally true of the first method, too.