Answer to Question #153677 in Differential Equations for RUPESH KUMAR

Question

Solve the following ordinary differential equations:

d²y/dx²+dy/dx+y=0

Answer 1

For solving equation : $y^{(2)}+y^{(1)}=-y$ .

The solution is $y=e^{q*x}$ .

By substituting such expression into the differential equation can be formulated as the equation on $q$ : $(q^{2} + q - 1)*e^{q*x}=0$ .

Where we substitute: $y^{(2)} = q^{2}*y$ and $y^{(1)} = q$ .

$e^{q*x} \neq 0$ as the exponential function will always be non-zero  for any finite $x$ so zeros must come from:

$q^2 + q + 1 =0$

solution of obtained equation is

$q_{+,-} = -\frac{1}{2} \pm \frac{i 3^{1/2}}{2}$

so solutions of the equation are

$y_{+,-} = C_{+,-}* exp(-\frac{1}{2} \pm \frac{i 3^{1/2}}{2})$

And the general solution of equation is

$y = C_{+}* exp(-\frac{1}{2} + \frac{i 3^{1/2}}{2}) + C_{-}* exp(-\frac{1}{2} - \frac{i 3^{1/2}}{2})$

also the general solution can be presented in the trigonometrical form, by substituting the formula into Euler's identity $e^{(a+i*b)}=e^a(\cos{b}+i*\sin{b})$ the general solution will be given by

$y = C_{1}* e^{-\frac{x}{2}}*\sin{ \frac{\sqrt{3}}{2}x} +  C_{2}* e^{-\frac{x}{2}}*\cos{ \frac{\sqrt{3}}{2}x}$