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This is actually a problem in algebra as shall be seen. I need to find the general solution for the following differential equation:

$$y''''+8y''+16y=0$$

The characteristic equation for this is:

$$\lambda^4+8\lambda^2+16=0$$

Factoring out gives us:

$$(\lambda^2+4)^2=(\lambda^2+4)(\lambda^2+4)=0$$

This generates a set of double complex conjugate roots $\lambda_{1,2}=\pm i2$ and $\lambda_{3,4}=\pm i2$

The general solution I get is:

$$y=A\cos(2x)+B\sin(2x)+xC\cos(2x)+xD\sin(2x)$$

Is this correct? If not please explain in detail where I went wrong. Thank you so much.

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  • $\begingroup$ Generally, we write $2i$ rather than $i2$. $\endgroup$ Jan 17, 2015 at 4:18
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    $\begingroup$ If you want to check your answer, you could always plug each of the four basis solutions into the equation. $\endgroup$
    – hasnohat
    Jan 17, 2015 at 4:23

1 Answer 1

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This solution is correct. Good job!

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  • $\begingroup$ Thank you. It's all I needed to know. :) $\endgroup$ Jan 17, 2015 at 5:59

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