Question

Consider the differential equation
y′′+αy′+βy=t+e^(4t).
Suppose the form of the particular solution to this differential equation as prescribed by the method of undetermined coefficients is
yp(t)=A1t^2+A0t+B0te^(4t).
Determine the constants α and β.

Answer 1

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Answer 2

Derivatives of the particular solution:
\[\begin{align*}
y_p(t)&=A_1t^2+A_0t+B_0te^{4t}\\\\
{y_p}'(t)&=2A_1t+A_0+B_0(1+4t)e^{4t}\\\\
{y_p}''(t)&=2A_1+B_0(8+16t)e^{4t}
\end{align*}\]
Subbing into the ODE and matching up coefficients gives the system
\[\begin{cases}
2A_1+\alpha A_0=0&\text{(constant)}\\
2\alpha A_1+\beta A_0=1&(t)\\
\beta A_1=0&(t^2)\\
(8+\alpha)B_0=1&(e^{4t})\\
(16+4\alpha+\beta)B_0=0&(te^{4t})
\end{cases}\]
I haven't taken the time to look at this much, but I'm not sure it's guaranteed that you can find numerical values for \(\alpha\) and \(\beta\) unless you're given values for \(A_1,A_0,B_0\)...

Answer 3

I looked it up online and found that \[\beta = 0\] I would like some help in just how to start solving this problem. I look at it and I don't even know where to start.

Answer 4

If we assume that \(A_0\neq0\), then the third equation of the system tells us that \(\beta=0\).

With this info you can solve for \(\alpha\) in the fifth equation:
\[(16+4\alpha)B_0=0~~\implies~~16+4\alpha=0\]
(provided that \(B_0\neq0\)).