Consider the second order initial value problem

y''+3y'+2y=0, y(0)=1,y'(0)=0

Write this as a system of two first order ordinary differential equations. Make sure you clearly define the dependent variables and state their initial conditions

Question

Consider the second order initial value problem

y''+3y'+2y=0,      y(0)=1,y'(0)=0

Write this as a system of two first order ordinary differential equations. Make sure you clearly define the dependent variables and state their initial conditions

dumbcow · Answer

do you have to make it a system?

this is homogeneous so just use characteristic equation

---> r^2 +3r+2 = 0

anonymous · Answer

To solve this, we write it as a quadratic: k^2+3k+2=0 which can be factorised to (k+2)(k+1) giving two roots: k1=-2 and k2=-1. We now get our linearly independent solutions y1=e^(-2x) and y2=e^(-x). Our solution y(x)=Ae^(-2x)+Be^(-x) for constants A and B which is what we need to find now using the given values.

y(x)=Ae^(-2x)+Be^(-x)
y(0)=1=A+B

y'(x)=-2Ae^(-2x)-Be^(-x)
y'(0)=0=-2A-B

We now have two equations for A and B that we can solve:
1=A+B 
=> B=1-A
0=-2A-B
0=-2A-(1-A)
A=-1 and hence B=2.

Sub these values back into our equation for y(x) and we get:

y(x)=-e^(-2x)+2e^(-x)

anonymous · Answer

If you let $u=y^{\prime}$, then it follows that $u^{\prime} = y^{\prime\prime}$.

Thus, $y^{\prime\prime}+3y^{\prime}+2y = 0\implies u^{\prime}+3u+2y = 0 \implies u^{\prime} = -3u-2y$  The initial condition $y^{\prime}(0) = 0$ is then equivalent to $u(0)=0$.

Therefore, the original equation can be written as the system of first order ODEs
$$\left\{\begin{aligned}y^{\prime} &= u\ u^{\prime} &= -3u-2y\end{aligned}\right.$$ with intial conditions $y(0)=1$ and $u(0)=0$.

I hope this makes sense! :-)