# y′′ − 2y′ − 3y = 3e^2t?

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find the general solution of the given differential equation.

### 1 Answer

• Start by solving for the solutions to the homogeneous problem:

y” – 2y’ -3y = 0

As usual for this class of problems, assume a solution of the form y = exp(kt), so y’ = kexp(kt) and y” = (k^2)exp(k*t). Plugging these into the differential equation yields the characteristic equation:

k^2 – 2k – 3 = 0

(k-3)(k+1) = 0

k = 3 and k = -1

The homogeneous solution is therefore:

y_h = aexp(3t) + bexp(-t)

where a and b are the constants of integration.

Now use the method of undetermined coefficients to solve for a particular solution. We seek a solution that is a linear combination of the forcing (nonhomogeneous) function and all of its derivatives. In this case, that means we are looking for a solution of the form y_p = c*exp(2t). This solution is linearly independent of the solutions to the homogeneous problem, so we don’t have to do anything “special”.

Differentiate this trial solution, plug the results into the original differential equation, and solve for the undetermined coefficient, “c”.

y’_p = 2cexp(2t)

y”_p = 4cexp(2t)

4cexp(2t) – 4cexp(2t) – 3cexp(2t) = 3*exp(2t)

-3c = 3

c = -1

so the particular solution is y_p = -exp(2t).

The complete solution is given by the sum of the homogeneous and particular solutions:

y(t) = a*exp(3t) + b*exp(-t) – exp(2t)

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