Explinations needed, second degree linear equations with a constant

[C.S]

Hello,

I am in need of explinations on how to resolve second degree linear equations with a constant.
I had a maths test last saturday and this was the last question :
Resolve y"+3y'+2=cos(t)
Having almost fallen into the trap of resolving y"+3y'+2y=cos(t), I passed the 2 over by cos(t) and tried to resolve y"+3y'=cos(t)-2, however I doubt that that's possible (maybe I'm wrong).
Anyhow I am looking for some help in resolving this equation because I have the same equation given as a test tomorrow morning.

Many thanks,

Callum

 

[daveb]

Do you know how to solve the homogeneous equation?

 

[C.S]

Yes I do, y₀(t)=C₁.e^-3t+C₂ with (C₁,C₂) →ℝ
I believe that's correct.

 

[daveb]

Not quite. If y = e^rt, then working out y' and y'' gives you e^rt(r²+3r+2) = 0, so the roots -1 amd -2 give you two solutions. Remember, there are n solutions to an nth order ODE.

Now you just need to find the particular solution. Have you learned that yet?

 

[C.S]

I thought ay"+by'+cy=0 gave us ar^2+br+c=0, in my case c=0.
I do know how to find particular solutions.
I'm going back into class, I will check back later. Thank you for your help.

 

[daveb]

Oops! You're right. I read it wrong as y'' + 3y' +2y. My bad.

 

[C.S]

Most people in my class did the same, any ideas though?

 

[daveb]

You have the right idea - just find the particular solution for the -2 (should be easy to see it's a polynomial in t) and the cosine term.

 

[lurflurf]

y"+3y'=-2+cos(t)
clearly
(-2+cos(t))'''+(-2+cos(t))'=0
so
y⁽⁵⁾+y⁽³⁾+3y=0
or in operator form
(D^2+3)y=-2+cos(t)
(D^5+4D^3+3D)y=D(D^2+3)(D^2+1)y=0
so the inhomogeneous problem reduces to a homogeneous problem.
A solution to the new problem will be a solution to the old one.

 

[Kurt Peek]

Hi,

So you want to solve the second-order differential equation $y''+3y'=\cos(t)-2$. The corresponding homogeneous equation is $y''+3y'=0$, the solution of which is $C_1+C_2\exp(-3t)$, where $C_1$ and $C_2$ are undetermined coefficients. The forcing function $\cos(t)-2$ can be expressed as the sum of two terms, $f_1(t)=\cos(t)$ and $f_2(t)=2$. By the superposition principle, the particular solution $y_p(t)$ can be expressed as the sum of the particular solutions $y_{p1}(t)$ and $y_{p2}(t)$ corresponding to $f_1(t)$ and $f_2(t)$, respectively. The individual particular solutions can be found using the method of undetermined coefficients.

Consider first $y_{p1}(t)$. We choose a trial solution $y_{t1}(t)=A_0\cos(t)+A_1\sin(t)$, where $A_0$ and $A_1$ are undetermined coefficients. Filling in $y_{t1}''+3y_{t1}'=\cos(t)$ and comparing terms leads to the pair of simultaneous equations $-A_0+3A_1=1$ and $-3A_0-A_1=0$, the solution of which is $A_0=-\frac{1}{10}$ and $A_1=\frac{3}{10}$. So $y_{p1}(t)=-\frac{1}{10}\cos(t)+\frac{3}{10}\sin(t)$.

Next, consdier $y_{p2}(t)$. The forcing function is a zeroth order polynomial. As explained by http://books.google.com/books?id=Pj...method of undetermined coefficients:&f=false", since the homogeneous equation lacks a damping term (i.e., one proportional to $y$), we should choose as our trial solution a polynomial with degree one order higher than the forcing function, so we choose $y_{t2}(t)=B_0+B_1t$, with $B_0$ and $B_1$ undetermined coefficients. Filling in $y_{t2}''+3y_{t2}'=-2$ then leads to $B_0=0$ and $B_1=-\frac{2}{3}$. So $y_{p2}(t)=-\frac{2}{3}t$.

Combining the above results, I obtain $$y(t)=y_h(t)+y_{p1}(t)+y_{p2}(t)= C_1+C_2\exp(-3t)-\frac{1}{10}\cos(t)+\frac{3}{10}\sin(t)-\frac{2}{3}t$$ as the final general solution of the differential equation. I verified this solution with Maple.

Hope this helps!

Cheers, Kurt

 

[C.S]

Thanks for all your help, the maths teacher said it was a last add in the test so that no one could get it all right, no wonder especially in the time given.
I will try to meditate all of that and see if I can do it again.