I have been working on finding the form of the particular solution for the following differential equations, using Paul's online notes as a reference, and every time I submit my answers, they are wrong. I'm supposed to use the method of undetermined coefficients, for equations of the form $$y''+p(t)y'+q(t)y=g(t)$$ Could someone point out what I'm doing wrong? Thank you.
For $y''-3y'-18y=x^2e^{-2x}$, we have $g(t)=x^2e^{-2x}$. For $x^2$, the guess is $Px^2+Qx+R$, and for $e^{-2x}$, it is $Ae^{-2x}$. Combining these, the form for the particular solution should be: $$y_p=e^{-2x}(Px^2+Qx+R)$$
For $y''+2y'+37y=e^{-x}\sin(6x)-6$, we have $g(t)=e^{-x}\sin(6x)-6$. Then, the guess for $\sin(6x)$ is $P\cos(6x)+Q\sin(6x)$; thus, combining it all: $$y_p=e^{-x}(P\cos(6x)+Q\sin(6x))+R$$
For $y''+36y=3+\sin(6x)$, using the same process, I found $$y_p=P\cos(6x)+Q\sin(6x)+R$$
I also have an equation of the form $y'''-y''+36y'-36y=(x^2+3)e^x$, and I have no idea how to start here, since according to Paul's online notes, this method is only valid for equations of the above form -- only up to second order. As I mentioned, the above answers are wrong for some reason; I am clueless as to why.
Edit: I have realized that the process can be adapted to higher-order ODEs, and I am working on this. This should not alter the validity of above three solutions, though.
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$\begingroup$Lets do the first one.
$$\tag 1 y''-3y'-18y=x^2e^{-2x}$$
The homogeneous solution is:
$$y_h(x) = c_1 e^{-3 x}+c_2 e^{6 x}$$
Next, lets do the calculations for the particular solution $y_p=e^{-2x}(px^2+qx+r)$.
We have:
- $y_p=e^{-2x}(px^2+qx+r)$
- $y'_p = e^{-2 x} (2 p x+q)-2 e^{-2 x} \left(p x^2+q x+r\right)$
- $y''_p = 4 e^{-2 x} \left(p x^2+q x+r\right)-4 e^{-2 x} (2 p x+q)+2 p e^{-2 x}$
Substituting into $(1)$, yields:
$(4 e^{-2 x} \left(p x^2+q x+r\right)-4 e^{-2 x} (2 p x+q)+2 p e^{-2 x}) - 3(e^{-2 x} (2 p x+q)-2 e^{-2 x} \left(p x^2+q x+r\right))-18(e^{-2x}(px^2+qx+r)) = x^2e^{-2x}$
This reduces to:
$$-8 p x^2-14 p x -8 q x +2 p-7 q-8 r = x^2$$
Equating like terms, we have:
- $-8 p = 1 \implies p = -\dfrac{1}{8}$
- $-14 p - 8 q = 0 \implies 8 q = \dfrac{14}{8} \implies q = \dfrac{14}{64} = \dfrac{7}{32}$
- $2 p -7q - 8 r = 0 \implies -\dfrac{1}{4} - \dfrac{49}{32} - 8 r = 0 \implies r = -\dfrac{57}{256} $
This gives us a final solution of:
$$y(x) = y_h(x) + y_p(x) = c_1 e^{-3 x}+c_2 e^{6 x} +e^{-2 x} \left(-\dfrac{57}{256}+ \dfrac{7}{32}x -\dfrac{1}{8}x^2\right)$$
The second and third are the same process, just algebra.
The last one is also doable using Undetermined Coefficients. We have a homogeneous of:
$$y_h(x) = c_1 \cos 6x + c_2 \sin 6x + c_3 e^x$$
We would choose (we are multiplying by the extra $x$ because we have $e^x$ in the right hand side and that matches what we got for the homogeneous solution):
$$y_p(x) = x(a e^x + b x e^x + c x^2 e^x)$$
$\endgroup$ $\begingroup$You forgot to consider cases where the RHS has the same form as the homogeneous solution. Take the third equation for example, since the characteristic polynomial is $$ r^2 + 36 = 0 $$
You get $r = \pm 6i$ so the homogeneous solution has the form $$ y_h = A_1\sin(6x) + B_1\cos(6x)$$
Since the in-homogeneous equation has $\sin(6x)$ in it, you have to add a factor of $x$ to your guess, to the particular solution should have the form
$$ y_p = Px\sin(6x) + Qx\cos(6x) + R $$
The second equation works the same way.
For the last equation, you can use the same method, since it's valid for equations of any order. I don't know why your notes say otherwise.
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