0
$\begingroup$

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?

$\endgroup$
1
  • $\begingroup$ 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 $\det(A)=\det\begin{bmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{bmatrix}=\begin{vmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{vmatrix}$ $\endgroup$ Oct 6, 2019 at 15:17

1 Answer 1

1
$\begingroup$

$$ v_2'=-\frac{3}{4}te^{3t}\implies v_2=-\frac{3}{4}(3te^{3t}-9e^{3t}) $$ 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, $$ v_2=-\frac{1}{4}\left(te^{3t}-\frac13e^{3t}\right) \implies y=-\frac34(t+1)e^{-t+3t}-\frac1{12}(3t-1)e^{3t-t}=-\left(t+\frac23\right)e^{2t}. $$

$\endgroup$
1
  • $\begingroup$ Thank you for pointing it out! $\endgroup$ Oct 7, 2019 at 12:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.