1. ## Ivp

Given the IVP $\displaystyle y'' - 4y = 0, y(0) = 2, y'(0) = \alpha$

1.) Find sol'n of the IVP in terms of $\displaystyle \alpha$

2.) Find $\displaystyle \alpha$ such that as $\displaystyle t \rightarrow \infty$ the sol'n approached zero.

2. First, we find the auxiliary equation by the quadratic.

$\displaystyle m^{2}-4=(x+2)(x-2)=0$

This gives us:

$\displaystyle C_{1}e^{2x}+C_{2}e^{-2x}$

Now, by using y(0)=2, we can find C1:

$\displaystyle C_{1}e^{2(0)}+C_{2}e^{-2(0)}=2$

$\displaystyle C_{1}=2-C_{2}$

But $\displaystyle y'=2C_{1}e^{2x}-2C_{2}e^{-2x}$

Then using the other IC, we can find C2:

$\displaystyle 2(2-C_{2})e^{2(0)}-2C_{2}e^{-2(0)}=a$

$\displaystyle C_{2}=1-\frac{1}{4}a$

Then $\displaystyle C_{1}=\frac{1}{4}a+1$

Therefore, the IVP in terms of a is:

$\displaystyle (1+\frac{1}{4}a)e^{2x}+(1-\frac{1}{4}a)e^{-2x}$

Now, find the limit for the last portion of the problem. It should be kind of easy to see what a is now.

$\displaystyle \lim_{x\rightarrow{\infty}}[(1+\frac{a}{4})e^{2x}]+\lim_{x\rightarrow{\infty}}[(1-\frac{a}{4})e^{-2x}]$

What is the limit on the right when x is unbounded?. What will give 0?

3. Regarding the limit... won't this only be 0 when alpha (or a in your case) = -4? That'll take care of the first part and the 2nd part