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I keep trying to solve this but I end up needing to do integration by parts like 3 or 4 times. My only question is, is that going to be the only way to do this? If so it will literally take me hours. I don't need a solution, just a confirmation that if the requirement is to use variation of parameters, then will I need to take integrals that need to be done by IBP?

Thank you and sorry if this is not a good question.

JustHeavy
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    See method of undetermined coefficients? I suppose you are doing derivative of $y$ with respect to $x$... It makes sense that if $y$ is just degree $1$ polynomial of $x$, $y''$ is nothing. – AdditIdent Oct 19 '18 at 08:30
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    You don't get anything worse than $\int x e^{\pm 5x} dx$, do you? Those shouldn't take you hours to do, it can be done in seconds! See here: https://math.stackexchange.com/questions/70974/lesser-known-integration-tricks/71195#71195 – Hans Lundmark Oct 19 '18 at 09:50

1 Answers1

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Hint.

This is a linear DE so it's solution can be written as

$$ y = y_h + y_p\\ y''_h-25y_h=0\\ y''_p-25y_p=x $$

The solution for $y_h = c_1 e^{5x}+c_2 e^{-5x}$.

Assuming now

$$ y_p = c_1(x)e^{5x}+c_2(x)e^{-5x} $$

substituting into the complete DE we have

$$ e^{5 x} \left(C_1'(x)+10 C_1(x)\right)+e^{-5 x} \left(C_2'(x)-10 C_2(x)\right)-x = 0 $$

with $C_1(x) = c_1'(x), C_2(x) = c_2'(x)$

Now choosing and solving

$$ C_1'(x)+10 C_1(x)=0\\ e^{-5 x} \left(C_2'(x)-10 C_2(x)\right)-x = 0 $$

we have

$$ C_1(x) = K_1e^{-10x} = c_1'(x)\\ C_2'(x)-10 C_2(x)=x e^{5x} $$

This last equation in $C_2(x)$ can be solved applying again the variation of constants method etc.

NOTE

$$ c_1'(x) = K_1 e^{-10x}\\ c_2'(x) = K_2e^{10x}-\left(\frac x5+\frac{1}{25}\right)e^{5x} $$

etc.

Cesareo
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