Question about Method of Undetermined Coefficients?

Question

Here's the problem: Consider the DE $y''−5y'+6y=e^t\cos(2t)+e^{2t}(3t+4)\sin(t).$ Determine a suitable form $Y (t)$ if the method of undetermined coefficient is to be used.

So I understand that I'm supposed to come up with a "guess" for $Y(t)$ which contains an undetermined coefficient. For instance, if it were simply $y''−5y'+6y=e^t,$ I would begin with $Y(t)=Ae^t.$ Then I would take the derivative of $Y(t)$ twice and plug that back in to the original DE.

I also realize that when $\cos/\sin$ are involved, you need to include both (like if it were $y''−5y'+6y=\cos(2t),\; Y(t)$ would be $A\cos(2t)+B\sin(2t)$). But since they're already both in $g(t),$ I don't need to do that, right?

Right now my initial guess for $Y(t)$ is $Ae^t\cos(2t)+Be^{2t}(3t+4)\sin(t)$... Is this correct?

Answer 1

try $$y_P=Ae^{t}\left(B\cos(2t)+C\sin(2t)+((Dt+E)\cos(t)+(Ft+G)\sin(t))e^t\right)$$

Answer 2

I'll highlight the two terms of the function in the right-hand side with coloring: $$y''−5y'+6y=\color{blue}{e^t\cos2t}+\color{purple}{e^{2t}(3t+4)\sin t} \tag{$*$}$$

Now for a particular solution:

I also realize that when cos/sin are involved, you need to include both (like if it were y''−5y'+6y=cos2t, Y(t) would be Acos2t+Bsin2t). But since they're already both in g(t), I don't need to do that, right?

You're right that sine and cosine both appear in the right-hand side, but multiplied with different exponential functions! To be on the safe side, it's worth considering the blue and purple terms in $(*)$ separately and carefully following the instructions about what particular solution(s) to suggest. Before you add them, you could check if you have redundant terms.

• For the blue part in $(*)$, you suggest a solution of the form: $$\color{blue}{y_{p_1} = e^t\left( A\cos 2t + B \sin 2t \right)}$$

• For the purple part in $(*)$, the complete form of the standard suggestion would be: $$\color{purple}{y_{p_2}= e^{2t}\left( Ct\sin t+D\sin t + Et\cos t+F\cos t\right)}$$

Because of linearity or the superposition principle, you can add your suggestions ($y_p = \color{blue}{y_{p_1}}+\color{purple}{y_{p_2}}$) and find the six undetermined coefficients $A,B,C,D,E,F$ by substitution.

Answer 3

The solutions for the homogeneous equation are $e^{2t}$ and $e^{3t}$.

The particular solution takes the form of $$e^{t} [Acos(2t) + B sin(2t)] +te^{2t}[Csin(t) +D cos(t)]$$

We find A, B and C,D by plugging the particular solution in the inhomogeneous equation.

Answer 4

It's better to substitute $y=e^tz(t)$ and simplify the given equation

$$y'=e^t(z'+z) \implies y''=e^t(z''+2z'+z)$$ The equation becomes simply $$z''-3z'+2z=\cos(2t)+e^t(3t+4)\sin(t)$$

For the $\cos(2t)$ part, you should choose $z_p=A\cos(2t)+B\sin(2t)$