Solve the initial value problem $$y' = \frac{-4x}{y}, \ y(0)= y_0,$$

and determine how the interval, in which the solution exists depends on the initial value $y_0$

My attempt:

Solving the initial value problem is straightforward and I get: $$y^2= -4x^2+y_0^2$$

However I don't understand the next part. The solution I have states that $y_0=0$ will only give a zero solution in the plane. and proceeds to get

$$\lvert x\rvert \lt \frac{\lvert y_0 \rvert}{2}$$

Can you help me understand how this works

  • $\begingroup$ $y^2$ cannot be negative (but if $x$ was increased beyond $|y_0|/2$ it would be). $\endgroup$ – almagest May 9 '16 at 13:13
  • $\begingroup$ I'm sorry I don't quite understand $\endgroup$ – THISISIT453 May 9 '16 at 13:20
  • $\begingroup$ You have $y^2=y_0^2-4x^2$, so if $|x|>\frac{|y_0|}{2}$, then $y^2<0$, which is impossible. $\endgroup$ – almagest May 9 '16 at 13:23

If we have $$y^\prime = - \frac{4 x}{y} \implies \frac{\mathrm{d}y}{\mathrm{d}x} = - \frac{4 x}{y}.$$

Then, we can write $$\int y \mathrm{d}y=\int-4x\mathrm{d}x$$ and $$\frac{y^2}{2} + C_y = -2x^2 + C_x $$ We say, $C_x - C_y = C$, $$y^2 = -4x^x + C$$ thus $$y = \pm \sqrt{C-4x^2}.$$ As we know $y(x=0) = y_0$, it implies $C=y_0^2$. Therefore, we write $$y(x) = \pm \sqrt{y_0^2-4x^2}.$$

Finally, the solution existes in the reals if if $4x^2 < y^2_0$, or, $x\in[-2y_0,+2y_0]$. We can say the both Domain and Image of $y(x)$ are functions of $y_0$, or $y(x):[-2y_0,+2y_0]\mapsto[0,y_0]$

  • $\begingroup$ $\pm y_0/2$, not $\pm 2y_0$. $\endgroup$ – tilper May 9 '16 at 16:25

$y^2 = -4x^2 + y_0^2$. The RHS must be positive or zero (i.e., it can't be negative) because it is equal to the square of something (the square of $y$).

So we must have $-4x^2 + y_0^2 \ge 0$. This means $x^2 \le \dfrac{y_0^2}{4}$.

Then take the square root of both sides to get $|x| \le \dfrac{|y_0|}{2}$.

In the answer they have strict inequality. This is likely to avoid trivial solutions. Because if $|x| = \dfrac{|y_0|}{2}$, then we have $y_0^2 = 4x^2$, which would give us $y^2 = -4x^2 + 4x^2 = 0$, which means $y$ is identically $0$. Therefore to avoid this trivial solution we must require that $|x| \ne \dfrac{|y_0|}{2}$, and so all together we have $|x| < \dfrac{|y_0|}{2}$.

Regarding the other part, let's suppose that $y_0 = 0$. Then we have $y^2 = -4x^2$, but since $y^2 \ge 0$ for all $y$ and $-4x^2 \le 0$ for all $x$, the equation $y^2 = -4x^2$ is only satisfied at the point $(0,0)$.


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