5
$\begingroup$

I have a question here from Differential Equations. We are learning how to solve second order nonhomogenous equations using the method of undetermined coefficients. The equation at hand is $$y''+4y=3\sin(2t)$$ I understand that the solution to the corresponding homogenous equation is $$y_c(t)=c_1\cos(2t)+c_2\sin(2t)$$ And I understand that to solve the equation using this method, I have to find an equation with "undetermined coefficients" that I then add to $y_c(t)$. But what form does this part of the equation take?

p.s. The answer in the back of the book is $y=2\cos2t-(1/8)\sin2t-(3/4)t\cos2t$. The initial conditions were $y(0)=2, y'(0)=-1$, but I don't need help with initial conditions.

$\endgroup$
2
$\begingroup$

Your problem is $(D^2+4)[y] = 3 \sin (2t)$. Operate by $D^2+4$ once again to find $(D^2+4)^2[y] = (D^2+4)(3\sin(2t)) =0$. Then, $$ y = c_1 \cos( 2t)+c_2\sin( 2t) + c_3t\cos (2t) + c_4t \sin (2t) $$ is the general solution. Identify the first two terms as the homogeneous solution to the given differential equation. Hence form: $$ y_p = t(A \cos (2t)+ \sin (2t)) $$ via the $c_3$ and $c_4$ terms. Next, $$ (D^2+4)[y_p] = 3 \sin (2t)$$ will yield equations to fix $A$, $B$. Finally, apply the initial conditions to fix $c_1,c_2$.

$\endgroup$
2
$\begingroup$

Since you don't want any solutions that are linearly independent to the homogenous solution, you guess a solution of the form $$y_p(t)=At\cos(2t)+Bt\sin(2t)$$ determine the coefficients, and by linearity of the differential operator, form the solution of the inhomogenous problem as $$y(t)=y_c(t)+y_p(t).$$

$\endgroup$

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.