Question

x''=4y+e^(t)
y''=4x-e^(t)
Solve the differential equations by systematic elimination

Answer 1

so these are parametric equations

Answer 2

\[\begin{cases}x''(t)=4y+e^t\\y''=4x-e^t\end{cases}\]
Solving for \(x\) in the second equation gives
\[x=\frac{y''-e^t}{4}\]
Differentiating twice, you have
\[x''=\frac{y^{(4)}-e^t}{4}\]
Substitute into the first equation:
\[\begin{align*}\frac{y^{(4)}-e^t}{4}&=4y+e^t\\\\
\frac{1}{4}y^{(4)}-4y&=\frac{5}{4}e^t
\end{align*}\]
The characteristic equation is
\[\frac{1}{4}r^4-4=0\]
which has four roots.
\[r^4-16=(r-2)(r+2)(r-2i)(r+2i)=0~~\implies~~r=\pm2,\pm2i\]
so your homogeneous solution is
\[y_h=C_1e^{2t}+C_2e^{-2t}+C_3\cos2t+C_4\sin2t\]
For the nonhomogeneous part, you can use the method of undetermined coefficients. As your guess function, use
\[y=Ae^t~~\implies~~y^{(n)}=Ae^t\]
\[\begin{align*}\frac{A}{4}e^t-4Ae^t&=\frac{5}{4}e^t\\\\
-\frac{15}{4}Ae^t&=\frac{5}{4}e^t\\\\
A&=-\frac{1}{3}
\end{align*}\]
and hence the general solution is
\[y(t)=y_h-\frac{1}{3}e^t\]

Answer 3

i am confused by the nonhomogenous part, how do you 'guess' the right function

Answer 4

It's the standard procedure of the method of undetermined coefficients.

If you have a second order ODE with constant coefficients, i.e. of the form
\[ay''+by'+cy=f(x)\]
you can check to see if a variant of \(f(x)\) fits as a solution for \(y\).

As a particular example, suppose \(f(x)\) is a polynomial. Then we can suppose \(y_p\), our guess solution, is also a polynomial, because derivatives of polynomials are also polynomials.

This method generally works if \(f(x)\) is a linear combination of sines, cosines, exponentials, and (I think, I could be wrong) what I might call "polynomial-periodic" functions, like \(x^2\sin2x\).