Find solutions y1 and y2 of the equation y"=0 that satisfy the initial conditions y1(xo)=1, y'1(xo)=0 and y2(xo)=0, y'2(xo)=1.

Question

idealist10 · Answer

r^2=0
r=0 (double root)
y=c1+c2x
Now what? @SithsAndGiggles

anonymous · Answer

What are these initial conditions? Just general $y(x_0)=0$ ?

anonymous · Answer

1, not 0*

idealist10 · Answer

The initial conditions are$$y _{1}(x _{0})=1, y'_{1}(x _{0})=0, y _{2}(x _{0})=0, y'_{2}(x _{0})=1$$

anonymous · Answer

Alright I was just wondering if you were given a particular value for $x_0$.

From your solution we know the two fundamental solutions. $y=C_1+C_2x$ admits $y_1=1$ and $y_2=x$ (or possibly the other way around, with $y_1=x$ and $y_2=1$ - we'll find out shortly).

With $y_1(x_0)=1$ and ${y_1}'(x_0)=0$, it's likely that $y_1$ is the constant solution, since
$$\begin{cases}y_1=C_1\{y_1}'=0\end{cases}\quad\text{and}\quad\begin{cases}y_1(x_0)=1\{y_1}'(x_0)=0\end{cases}~~\implies~~C_1=1$$
Similarly, if $y_2=C_2x$, then
$$\begin{cases}y_2=C_2x\{y_2}'=C_2\end{cases}\quad\text{and}\quad\begin{cases}y_2(x_0)=0\{y_2}'(x_0)=1\end{cases}~~\implies~~C_2=1$$
(and also that $x_0=0$).

idealist10 · Answer

The answers are $$y _{1}=1, y _{2}=x-x _{0}$$

idealist10 · Answer

How did you find C1=1 and C2=1 and also that xo=0?

anonymous · Answer

Ignore everything past "Similarly,..." That's wrong. I thought the work suggested that $x_0=0$, but that's not the case, and it is in fact a general constant.

If you can accept that the fundamental solutions are $y_1=1$ and $y_2=x$, finding the values of $C_1$ and $C_2$ is a matter of solving these simple IVPs.

We have the ODE $y''=0$ with initial conditions $y_1(x_0)=1$ and ${y_1}'(x_0)=0$. You know the general solution is $y=C_1+C_2x=C_1y_1+C_2y_2$. The principle of superposition allows us to consider one solution at a time.

The first initial condition tells us that
$$y_1(x_0)=1~~\implies~~C_1=1$$
The second says
$${y_1}'(x_0)=0,\quad\text{which doesn't give us any more info}$$

anonymous · Answer

Now consider the second solution, $y_2=C_2x$, and its initial conditions: $y_2(x_0)=0$ and ${y_2}'(x_0)=1$. We then have
$$y_2(x_0)=C_2x_0=0~~\implies~~C_2=0\text{ or }x_0=0$$
Assuming $C_2\neq0$, this implies the constant $x_0$ is zero, but let's ignore this for the time being.

We also have
$${y_2}'(x_0)=C_2~~\implies~~C_2=1$$
and so $y_2=x$. I'm not quite sure how they're getting $x-x_0$...

idealist10 · Answer

So what's the correct answer for this problem?

idealist10 · Answer

@SithsAndGiggles

idealist10 · Answer

@SithsAndGiggles

anonymous · Answer

I haven't changed my mind, I'm still not sure how the answer should be $x-x_0$. Have you tried asking your instructor?

idealist10 · Answer

Yes, but he's not sure either.