Question

Consider the differential equation ay'' + by' + cy = 0 where the characteristic equation a*r^2 + b*r + c = 0 has complex roots λ ± iμ.

I need to show that u(t) = e^(λt) cos(μt) and v(t) = e^(λt) sin(μt) are solutions by substituting them for y in the differential equation. When I do this, though, I'm having trouble manipulating the equation to make it equal to zero.

Does anyone have any suggestions?

Answer 1

\[ay''+by'+cy=0\]
\[\begin{align*}y=u(t)&=e^{\lambda t}\cos\mu t\\
y'&=\lambda e^{\lambda t}\cos\mu t-\mu e^{\lambda t}\sin\mu t\\
&=e^{\lambda t}\left(\lambda\cos\mu t-\mu\sin\mu t\right)\\
y''&=\lambda e^{\lambda t}\left(\lambda\cos\mu t-\mu\sin\mu t\right)+e^{\lambda t}\left(-\mu\lambda\sin\mu t-\mu^2\cos\mu t\right)\\
&=e^{\lambda t}\left(\lambda^2\cos\mu t-\lambda\mu\sin\mu t-\lambda\mu\sin\mu t-\mu^2\cos\mu t\right)\\
&=e^{\lambda t}\left(\left(\lambda^2-\mu^2\right)\cos\mu t-2\lambda\mu\sin\mu t\right)
\end{align*}\]
\[ae^{\lambda t}\left(\left(\lambda^2-\mu^2\right)\cos\mu t-2\lambda\mu\sin\mu t\right)+be^{\lambda t}\left(\lambda\cos\mu t-\mu\sin\mu t\right)\\
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~+ce^{\lambda t}\cos\mu t=0\\
e^{\lambda t}\left[a\left(\lambda^2-\mu^2\right)\cos\mu t-2a\lambda\mu\sin\mu t+b\lambda\cos\mu t-b\mu\sin\mu t+c\cos\mu t\right]=0\\
a\left(\lambda^2-\mu^2\right)\cos\mu t-2a\lambda\mu\sin\mu t+b\lambda\cos\mu t-b\mu\sin\mu t+c\cos\mu t=0\\
\left(a\left(\lambda^2-\mu^2\right)+b\lambda+c\right)\cos\mu t+\left(-2a\lambda\mu-b\mu\right)\sin\mu t=0\]
Match up coefficients of common terms. There are no trig expressions on the right side, so you know that the coefficients on the left must equal zero:
\[\begin{cases}
a\left(\lambda^2-\mu^2\right)+b\lambda+c=0\\
-2a\lambda\mu-b\mu=0
\end{cases}\]
I think you'll have to assume \(\lambda,\mu\not=0\) to solve for \(a,b,c\). Besides, if \(\mu\) or \(\lambda\) were zero, you'd have the trivial solutions \(e^{\lambda t}\) and 0.