Differential Equation Help

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Homework Statement


Find constants A, B, and C such that the function y = Ax^2+Bx+C satisfies the differential equation y''+y'-2y=x^2

2. The attempt at a solution

d/dx (y) = d/dx (Ax^2+Bx+C) = 2Ax+B
y' = 2Ax+B

d/dx (y') = d/dx (2Ax+B) = 2A
y'' = 2A

Now subbing back into the differential equation given:
(2A) + (2Ax+B) - 2y = x^2
2A + (2Ax+B) - 2(Ax^2+Bx+C) = x^2
2A + 2Ax + B - 2Ax^2 - 2Bx - 2C = x^2

Moving the left hand side around:
(2A-2C+B) + (2A-2B)x - 2Ax^2 = x^2

However, now I have to solve for the constants and I'm not exactly sure how to figure that out. Any help would be great! Thanks guys :D
 
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Answers and Replies

  • #2
2,063
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Well, in the final equation, you have an x^2 term, an x term and a constant on the left, while you have only an x^2 term on the right. Now, based on this, what can you conclude about the coefficients on the left?
 
  • #3
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Well the coefficients on the left hand side if we were to rearrange it to
-2Ax^2 + (2A-2B)x + (2A-2C+B)
would be -2, 1, 1 which in a way is kind of similar to the equation:
-2x^2 + x+ 1 which is a polynomial. However, I'm not too sure if I should be moving the x^2 term from the right hand side over to the left hand side and solving for 0. Haha I'm sorry if I didn't really answer your question.
 
  • #4
2,063
2
Okay, let me put in a different way...

Suppose ax^2+bx+1 = 3x^2+cx+d (a,b,c,d constants)

What can you say about a, b, c and d?
 
  • #5
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Hmmm I must not be seeing the logic behind this at all, and usually I would. Maybe I'm just having an off day or something. 12 years of schooling and 1 year of University is failing me today. The only thing I can think of would be that you are moving the right hand side over and subtracting it to be:
ax^2+bx+1 - 3x^2 - cx - d = 0
(a-3)x^2 + (b-c)x - d + 1 = 0

I feel really silly right now for not understanding because I know it is more likely really simple because the same principle for what you are asking will apply to my above question.
 
  • #6
D H
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The way to solve a problem like

[tex]a_nx^n + \cdots + a_1x + a_0 = A_nx^n + \cdots + A_1x + A_0[/tex]

is to split this one equation into n equations by grouping like powers of x:

[tex]a_nx^n = A_nx^n[/tex]
[tex]a_1x = A_1x[/tex]
[tex]a_0 = a_0[/tex]

from which [itex]a_k = A_k[/itex].

How does this work? Evaluate the expression at x=0. The constant terms are all that are left; they must be equal. The higher-order powers must be equal as well (prove recursively). You don't need the proof to solve the problem. Simply equate like terms.
 
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  • #7
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OHHH alright! Thank you so much both of you. That last explanation I think helped the most. I knew it could not be a hardship to figure it out but I made it seem harder than it should be. The keywords that helped me the most was "equate like terms" which I know is what neutrino was trying to state too.

Going back to my question, I'm finding that A = -1/2, B = -1/2 and C = -3/4 I believe. I'm hoping that is right anyway. :D
 
  • #8
D H
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OHHH alright! Thank you so much both of you. That last explanation I think helped the most.
Thanks.

Going back to my question, I'm finding that A = -1/2, B = -1/2 and C = -3/4 I believe. I'm hoping that is right anyway. :D
Bingo!
 
  • #9
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Excellent, yeah I just went back and subbed into my formula the answers I just got:
- 2Ax^2 + (2A-2B)x + (2A-2C+B) = x^2

Which ends up equaling x^2 on both sides, so the identity must be true.
 

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