Consider the IVP
$y''-2y'+26y=0$, $y(0)=1$, $y'(0)=2$.
From the characteristic equation $m^2-2m+26=0$, i found the roots as $m_1=1-5i$ and $m_2=1+5i$. Then when i use the basis solutions $y_1(x)=e^x\cos(5x)$ and $y_2(x)=e^x\sin(5x)$, the solution of IVP is of the form
$y(x)=e^x(\cos(5x)+\frac{1}{5}\sin(x))$
But when i use the basis solutions $e^{(1-5i)x}$ and $e^{(1+5i)x}$, then the solution of IVP is of the form
$y(x)=e^x\Big(\cos(x)+(5+2i)\sin(x)\Big)$.
I know that the solution of this IVP is unique but these solution are different. The question is how can i show that these solution are equivalent. I have really no idea. Any help will be appreciated.
Thanks in advance!...
$\endgroup$ 11 Answer
$\begingroup$I would say
$y=C_1e^{(1+5i)x}+C_2e^{(1-5i)x}=e^x(C_1e^{5ix}+C_2e^{-5ix}=e^x((C_1+C_2)\cos 5x+i(C_1-C_2)\sin 5x)=e^x(A\cos 5x+B\sin 5x)$
And according to the initial conditions $y=e^x(\cos 5x-\frac{1}{5}\sin 5x)$
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