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How to go about solving this:

$$y'' + 4y' + 5 = 0$$

$$y = Ae^{px} + Be^{qx}$$

I know the following:

p and q are solutions to the characteristic equation $am^2 + bm + c = 0$

So in this case $m^2 + 4m + 5 = 0$

However I do not know what to do after this.

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your equation is $y''+4y'+5=0$? – user52188 Jan 9 '13 at 4:57
yes, thanks, post edited. – Dreamer78692 Jan 9 '13 at 4:59
do you want solutions real or complex? – user52188 Jan 9 '13 at 5:04
I needed solution in the real domain, but I guess thats not possible in this case. – Dreamer78692 Jan 9 '13 at 5:11
up vote 5 down vote accepted

If you want all solution real:

$m^2+4m+5=0$ $\implies$ $m=-2\pm i$.

Then $e^{(-2+i)t}=e^{-2t}(\cos{t}+i\sin{t}) $ and $e^{(-2-i)t}=e^{-2t}(\cos{t}-i\sin{t})$ are two solutions.

Clearly linear combinations these solutions still are solutions. Then




We have $\phi$ and $\xi$ are two solutions linearly independent. Therefore

all real solution are $C_1e^{-2t}\cos{t}+C_2e^{-2t}\sin{t}$ for $C_1,C_2\in \mathbb R$

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@Dreamer78692 it's possible find real solution – user52188 Jan 9 '13 at 5:23

So, $$m=\frac{-4\pm\sqrt{4^2-4\cdot1\cdot5}}2=-2\pm i$$

Then as the roots are unequal, $$y=Ae^{(-2+i)x}+Be^{(-2-i)x}$$ where $A,B$ are arbitrary constants.

$$y=e^{-2x}(Ae^{ix}+Be^{-ix})$$ $$=e^{-2x}\{(A+B)\cos x+i(A-B)\sin x\}$$ using Euler's identity.

or, $$y=e^{-2x}(c_1\cos x+c_2\sin x)$$ where $c_1=A+B,c_2=i(A-B)$ are arbitrary constants.

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@Dreamer78692, we are free to keep $c_1,c_2$ real or complex whatever we like to. – lab bhattacharjee Jan 9 '13 at 5:29
Thanks man, i understand now. – Dreamer78692 Jan 9 '13 at 5:39
@Dreamer78692, welcome. Just a query, why you are interested in real solutions only? – lab bhattacharjee Jan 9 '13 at 5:41
The question asked to find all real solutions, Is should have put that in my post, sorry about that. – Dreamer78692 Jan 9 '13 at 5:44

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