Question

The indicated function y1(x) is a solution of the given differential equation. Use reduction of order or formula (5), as instructed, to ?nd a second solution y2(x).
y``+2y`+y=0

Answer 1

Answer:

Step-by-step explanation:

We will use the reduction of order to solve this equation. At first, we need a solution of the homogeneus solution.

Consider the equation $y''+2y'+y=0$ We will assume that the solution is of the form $y=Ae^{rx}$. If we plug this in the equation, we get

$Ae^{rx}(r^2+2r+1)=0$

Since the exponential function is a positive function, and A should be different to zero to have non trivial solutions, we get

$r^2+2r+1=0$

By using the quadratic formula, we get the solutions

$r= \frac{-2\pm \sqrt[]{4-4}}{2}=-1$

So one solution of the homogeneus equation is of the form $y=Ae^{-x}$. To use the reduction of order assume that

$y = v(x)y_h$

where $y_h = Ae^{-x}$. We calculate the derivatives and plug it in the equation

$y' = v'y_h+y_h'v$

$y'' = v''y_h+v'y_h'+y_h'v'+y_h''v = v''y_h+2v'y_h+y_h''v$

$(v''y_h+2v'y_h'+y_h''v)+2(v'y_h+y_h'v)+vy_h = 0$

If we rearrange the equation we get

$v''y_h+(2y_h'+2y_h)v'+v(y_h''+2y_h'+y_h)=0$

Since $y_h$ is a solution of the homogeneus equation we get

$v''y_h+(2y_h'+2y_h)v'=0$

If we take w = v', then w' = v''. So, in this case the equation becomes

$w'y_h+(2y_h'+2y_h)w=0$

Note that $y_h' = -1y_h$ so

$w'y_h=0$. Since $y_h$ cannot be zero, this implies

w' =0. Then, w = K (a constant). Then v' = K. So v=Kx+D where D is a constant.

So, we get that the general solution is

$y = vAe^{-x} = (Kx+D)Ae^{-x} = Cxe^{-x} + Fe^{-x}$ where C, F are constants.