I am trying to solve for the particular solution of the following Differential Equation:

$$y'' - 2y' - 3y = 3te^{2t}$$

using the method of Variation of Parameters.

I have already solved this DE by using Undetermined Coefficients. However, I do not get the same particular solution as MVP.

Here's my work:

$$y_1 = e^{3t}$$ $$y_2=e^{-t}$$ $$g(t)=3te^{2t}$$

Solving for the Wronskian:

$$W=-4e^{2t}$$ (I don't know how to format the determinant solution here)

The particular solution is given by the following formula:

$$y_p=v_1y_1 + v_2y_2$$

where $v_i'=\frac{W_i}{W}$

Solving, I got:
$$W_1=-3te^t$$ $$W_2=3te^{5t}$$ $$v_1'=\frac{3}{4}te^{-t}\implies v_1=\frac{3}{4}(-te^{-t}-e^{-t})$$ $$v_2'=-\frac{3}{4}te^{3t}\implies v_2=-\frac{3}{4}(3te^{3t}-9e^{3t})$$ Hence, we have the particular solution:
$$y_p=-3te^{2t}+6e^{2t}$$ which is different to the one I got using Undetermined Coefficients:
$$y_p=-te^{2t}-\frac{2}{3}e^{2t}$$ which I am pretty confident to say that it is correct.

I don't know where I did my solution wrong. Can anyone help me out here?

  • $\begingroup$ As to your formatting question, $\det(A)=\det\begin{bmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{bmatrix}=\begin{vmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{vmatrix}$ gives $\det(A)=\det\begin{bmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{bmatrix}=\begin{vmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{vmatrix}$ $\endgroup$ Commented Oct 6, 2019 at 15:17

1 Answer 1


$$ v_2'=-\frac{3}{4}te^{3t}\implies v_2=-\frac{3}{4}(3te^{3t}-9e^{3t}) $$ seems suspicious. Check by computing the derivative $$ v_2'=-\frac{3}{4}(3e^{3t}+9te^{3t}-27e^{3t})=-\frac{3}{4}(9te^{3t}-24e^{3t}) $$ which in both terms is not what the original derivative was. Correctly integrated it gives the correct solution, $$ v_2=-\frac{1}{4}\left(te^{3t}-\frac13e^{3t}\right) \implies y=-\frac34(t+1)e^{-t+3t}-\frac1{12}(3t-1)e^{3t-t}=-\left(t+\frac23\right)e^{2t}. $$

  • $\begingroup$ Thank you for pointing it out! $\endgroup$ Commented Oct 7, 2019 at 12:15

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