Initial value problem, show y(x) = 1-x is a solution

Question

I'm not really sure on how to go about solving this problem...

Consider the initial value problem.

$$y' (x) = \frac{1}{2}(-x + \sqrt{x^2 + 4y})$$ $$y(2) = -1$$

$a)$ Show that $y(x) = 1-x$ and $y(x) = -\frac{x^2}{4}$ are two solutions to the above IVP.

Show they both satisfy the IC and then differentiate each individually to show they satisfy the ODE. — Matthew Cassell, Feb 12 '17 at 09:29
Since you're not asked to solve the ODE itself, just take each of the given $y(x)$ and show that it satisfies the two given conditions. — George Law, Feb 12 '17 at 09:31
Another related question: https://math.stackexchange.com/q/1643067/115115. — Lutz Lehmann, Feb 12 '17 at 12:15

score 0 · Answer 1 · answered Feb 12 '17 at 09:29

0

You should just compute the derivative wrt $x$ of the suggested $y(x)$ and then plug into the first equation. Finally plugin the value $x=2$ and verify that the first equation holds.

Do the same for the other function $y(x)$.

answered Feb 12 '17 at 09:29

Maczinga

1,155

score 0 · Accepted Answer · answered Feb 12 '17 at 09:37

Consider $f(x)=1-x$; then $f'(x)=-1$, so $f'(2)=-1$. Also $$ \frac{1}{2}\bigl(-x+\sqrt{x^2+4f(x)}\,\bigr)= \frac{1}{2}\bigl(-x+\sqrt{x^2-4x+4}\,\bigr)= \frac{1}{2}\bigl(-x+|x-2|\,\bigr)= \begin{cases} -1 & \text{if $x\ge2$}\\ 1-x & \text{if $x<2$} \end{cases} $$ Thus the function $f$ is a solution of the problem only on the interval $[2,\infty)$. On $(-\infty,2]$ we have $$ \frac{1}{2}\bigl(-x+\sqrt{x^2+4f(x)}\,\bigr)=1-x\ne f'(x) $$

On the other hand, if $g(x)=-x^2/4$, $g'(x)=-x/2$ and $$ \frac{1}{2}\bigl(-x+\sqrt{x^2+4g(x)}\,\bigr)=-\frac{x}{2} $$ Thus the function $g$ is a solution over $\mathbb{R}$.

If you want a solution in a complete neighborhood of $2$, you have to choose the second one.

Initial value problem, show y(x) = 1-x is a solution

2 Answers2