Question

Verify that the given functions form the fundamental set of solution of the

differential equation on the given indicated interval.

y"-2y' + 5y = 0

y= c₁e^x cos 2x + c2₂e^x sin 2x, (-infinity, +infinity)

Answer 1

The characteristic equation $k^2-2k+5=0$ of the differential equation $y''-2y' + 5y = 0$ is equivalent to $(k-1)^2=-4,$ and hence has the roots $k_1=1+2i$ and $k_2=1-2i.$

Let us show that $e^x\cos 2x$ and $e^x\sin 2x$ form the fundamental set of solution.

Consider the equality $ae^x\cos 2x+be^x\sin 2x=0.$ If $x=0,$ then $ae^0\cos 0+be^0\sin 0=0$ implies $a=0.$ If $x=\frac{\pi}4,$ then $ae^0\cos \frac{\pi}2+be^0\sin \frac{\pi}2=0$ implies $b=0.$ Therefore, $e^x\cos 2x$ and $e^x\sin 2x$ are linearly independent, and hence they form the fundamental set of solution.

We conclude that the general solution is $y=C_1e^x\cos 2x+C_2e^x\sin 2x.$