Solve the Differential Equation y'' + y' - 6y = e^t

To solve this differential equation, we first find the characteristic equation:

r^2 + r - 6 = 0

This factors as:

(r + 3)(r - 2) = 0

So the roots are r = -3 and r = 2. Therefore, the general solution to the homogeneous equation (y'' + y' - 6y = 0) is:

y_h(t) = c1e^-3t + c2e^2t

To find a particular solution to the non-homogeneous equation (y'' + y' - 6y = e^t), we can use the method of undetermined coefficients. Since e^t is already a solution to the homogeneous equation, we try a particular solution of the form:

y_p(t) = Ate^t

Taking the first and second derivatives of y_p, we get:

y'_p(t) = Ae^t + Ate^t

y''_p(t) = 2Ae^t + Ate^t

Substituting these into the original equation, we get:

(2A + Ae^t + Ate^t) + (Ae^t + Ate^t) - 6Ate^t = e^t

Simplifying and collecting terms, we get:

(3A + 1)te^t = e^t

Therefore, A = 1/3. So the particular solution is:

y_p(t) = (1/3)te^t

The general solution to the non-homogeneous equation is then:

y(t) = y_h(t) + y_p(t) = c1e^-3t + c2e^2t + (1/3)te^t

where c1 and c2 are constants that depend on the initial conditions.