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# Solve the Differential Equation y"-y'-2y=3x+4? watch

1. I must find the general solution to the second order differential equation:
y"-y'-2y=3x+4

I know the solution must be:
y(x) = C1*e^(2x)+C2*e^(-x)-3x/2-5/4

I'm having trouble finding the particular solution.
I tried this problem as follows:

For the complimentary solution:
r^2-r-2=0
(r-2)(r+1)=0
r=2,-1
yc(x)=C1*e^(2x)+C2*e^(-x)

For the particular test solution:
yt(x) = Ax^2+Bx+C
yt'(x) = 2Ax+B
yt"(x) = 2A

yt(x) = (Ax^2+Bx+C)-(2Ax+B)-2(2A)
yt(x) = Ax^2+Bx+C-2Ax-B-4A
^^^ Would I use parenthesis for each term and therefore distribute the negative signs, or should I just plug them in?
Equations:
x^2: A = 0
x: B-2A = 1
constants: C-B-4A = 1
^^^ Equals number of like terms in original DE on RHS?
So, A=0, B=1, C=2

yp(x) = x+2

y(x) = yc(x) + yp(x)
y(x) = C1*e^(2x)+C2*e^(-x)+x+2

My complimentary solution is correct, but where am I going wrong with my particular solution?
2. 2A - (2Ax+B) - 2(Ax^2+Bx+C) = 3x+4 ...you mean
3. Well, yes, but don't you set up the system of equations based on the number of like terms on the RHS?
4. Oh, I see what happened in the class example, I missed the simplifications. I'll see if I can try this again.
5. Setting up my equations as follows:
x^2: -A = 0
x: -2A-B = 3
const: 2A-B-C = 4
A=0, B=-3, C=-1

Which still isn't right...
6. -2a = 0
-2a-2b = 3
2a-b-2c = 4
7. Just so you know in future, because the RHS is linear, there is no need to try a quadratic particular integral, you just need to try a linear one.
8. Oh, duh, that time I forgot to distribute the 2 >.<

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