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Find a second order differential equation so that $$y=C_1e^{-3x}\cos(4x)+C_2e^{-3x}\sin(4x)+4e^{3x}$$ solves the differential equation for any choice of $C_1$and $C_2$.

The answer should be in the form of $ay''+by'+cy=f$

Here's my work: $y=C_1e^{-3x}\cos(4x)+C_2e^{-3x}\sin(4x)$ is the solution of the homogeneous equation and $y=4e^{3x}$ is the particular solution. But how do I proceed from here to figure out the second order ODEs?

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You have to "solve" the equation backwards. As you pointed, there are solutions to the homogeneous equation and a particular solution. To get the homogeneous equations, we notice that the roots of the characteristic polynomial equation are $-3+4i$ and $-3-4i$. and the characteristic polynomial equations is therefore $$(x+3-4i)(x+3+4i)=x^2+6x+25$$ So, the homogeneous equation is $$y''+6y'+25y=0$$ To get the full equation, we substitute the particualr solution $$(4e^{3x})''+6(4e^{3x})'+25(4e^{3x})=208e^{3x}$$ We get therefore, the equation $$y''+6y'+25y=208e^{3x}$$

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Hints:

  • We have complex conjugate roots that are $3 \pm 4 i$. What equation gives those roots? This defines the homogeneous equation $ay'' + by' + cy = 0$ .
  • We know the particular solution result and using the homogeneous result we just derived above, we can find what the constant for the particular solution is for the DEQ. That is, substitute $y = 4 e^{3x}$ into the homogeneous result.

Spoiler

$y'' + 6 y' + 25 y = 208 e^{3x}$

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  • $\begingroup$ The "spoiler" is getting to be your trademark! +1 $\endgroup$ – Namaste Oct 18 '13 at 0:02

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