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Solve $y''+4y'+5y=0$ $y(0)=1$ $y'(0)=0$

My solution so far

$x^2+4x+5x=0$

Solving the quadratic yields $x\in\{-2+i,-2-i\}$

$y_1=e^{-2x}(\cos(x)+i\sin(x))$

$y_1=e^{-2x}(\cos(x)-i\sin(x))$ .

Combining the two .

$y(x)=e^{-2x}(c_1\cos(x)+c_2\sin(x))$ .

Here, I don't know how to solve for $c_1, c_2$ given $y(0)=1$ $y'(0)=0$

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$$ y(0)=e^{-2\times 0}(c_1\cos(0)+c_2\sin(0)) $$

$$ y(0)=c_1 $$

One can see that $y(0)=1$ only if $c_1=1$.

You now need to take your general solution, differentiate plug in $x=0$ and demand that that is equal to $0$.

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