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14.8: A First Order Differential Equation

Example \(\PageIndex{1}\)

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


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:

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


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

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.