Find that solution satisfying $ϕ$ satisfying $ϕ(0) = 0$, $ϕ'(0) = 1$, $ϕ''(0)=0$ for $y'''-4y'=0$.

Question

Problem:

Consider the equation $y'''- 4y' = 0$

Find that solution satisfying $ϕ(0) = 0$, $ϕ'(0) = 1$, $ϕ''(0)=0$.

My solution:

The characteristic polynomial of $y'''- 4y' = 0$ is $r^3 - 4r = 0$.

Then, $r(r^2 - 4) = 0$ → $r(r-2)(r+2) = 0$ → $r = 0,r = 2,r = -2$.

Therefore, $ϕ(x) = c_1 + c_2e^{2x} + c_3e^{-2x}$

As given,

i. $ϕ(0) = c_1 + c_2 + c_3 = 0$

ii. $ϕ'(x) = 2c_2e^{2x} - 2c_3e^{-2x}$ then, $ϕ'(0) = 2c_2 - 2c_3 = 1$

iii. $ϕ''(x) = 4c_2e^{2x} + 4c_3e^{-2x}$ then, $ϕ''(0) = 4c_2 + 4c_3 = 0$. Therefore, $c_2 = -c_3$. Substituting $c_2$ in to ii yeilds $c_3 = \frac{-1}{4}$. Then $c_2 = \frac{1}{4}$. Substituting $c_2, c_3$ to i yields $c_1 = 0$.

So, the solution satisfying $ϕ(0) = 0$, $ϕ'(0) = 1$, $ϕ''(0)=0$ is $ϕ(x) = \frac{1}{4}e^{2x} + \frac{-1}{4}e^{-2x}$

But book says $ϕ(x) = \frac{sinh(2x)}{2}$. Can someone please verify my answer? Thanks!

Answer 1

Your solution is correct. Note that by definition: $$\sinh(x)\equiv \frac{e^x-e^{-x}}{2}$$ Thus: $$\phi(x)=\frac{1}{4}e^{2x}-\frac{1}{4}e^{-2x}=\frac{e^{2x}-e^{-2x}}{4}=\frac{\sinh(2x)}{2}$$