Help finding the answer to DE using method of undetermined coefficients

lilmul123
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Homework Statement



y'' - 5y' = 2x^3 - 4x^2 -x + 6




The Attempt at a Solution



I first found the m's on the left hand side. They are 0 and 5. I then found the m's on the right hand side. They are 0, 0, 0, and 0. I then wrote my interim equation:

y = C1 + C2e^5x + Ax^4 + Bx^3 + Cx^2 + Dx. Because there already is a 0 on the left hand size for m, then there are x's for each part of the equation on the right side.

This is where I'm confused. I have tried to find the coefficients by finding the first and second derivatives of Yp, but I get very odd numbers when I solve for the variables(B = 14/75 and C = 5.324 for example). Am I on the right track or did I screw up somewhere? We have never been presented with irrational coefficients before (as is the case with C), so I'm thinking I did something wrong somewhere...
 
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It sounds like you're on the right track. I think you're just making algebraic errors when solving for A, B, C, and D, so double-check your work for mistakes. If you post your actual work in solving for them, others and I can see if we spot any mistakes as well.
 
D'oh. You were right. Thank you!
 
I don't see anything wrong, but don't use approximations, since that will foul things up. Instead of 5.324, use the exact value.
 
There were no approximations, my algebra was just poor.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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