# What is the solution of the differential equation? : ydy/dx = (9-4y^2)^(1/2) where x=0 when y=0

Sep 17, 2017

$y = \pm \sqrt{6 x - 4 {x}^{2}}$

#### Explanation:

We have:

$y \frac{\mathrm{dy}}{\mathrm{dx}} = {\left(9 - 4 {y}^{2}\right)}^{\frac{1}{2}}$ ..... [A}

Which is a First Order separable (non-linear) Differential Equation, so we can rearrange [A] to get:

$\frac{y}{\sqrt{9 - 4 {y}^{2}}} \frac{\mathrm{dy}}{\mathrm{dx}} = 1$

And we can now "seperate the variables as follows":

$\int \setminus \frac{y}{\sqrt{9 - 4 {y}^{2}}} \setminus \mathrm{dy} = \int \setminus \mathrm{dx}$ ..... [B]

Here the RHS is trivial to integrate, and we will need a substitution for the LHS integral:

$I = \int \setminus \frac{y}{\sqrt{9 - 4 {y}^{2}}} \setminus \mathrm{dy}$

Let us perform a substitution, and try:

$u = 4 {y}^{2} \implies \frac{\mathrm{du}}{\mathrm{dy}} = 8 y$

Then substituting into the integral, we get:

$I = \frac{1}{8} \setminus \int \setminus \frac{8 y}{\sqrt{9 - 4 {y}^{2}}} \setminus \mathrm{dy}$
$\setminus \setminus = \frac{1}{8} \setminus \int \setminus \frac{1}{\sqrt{9 - u}} \setminus \mathrm{du}$

And we now directly integrate this, giving:

$I = \frac{1}{8} \left(- 2 \sqrt{9 - u}\right)$
$\setminus \setminus = - \frac{1}{4} \sqrt{9 - u}$

And reversing the substitution we get:

$I = - \frac{1}{4} \sqrt{9 - 4 {y}^{2}}$

Using this result, we can now integrate our earlier result [B] to get the General Solution:

$- \frac{1}{4} \sqrt{9 - 4 {y}^{2}} = x + C$

We can apply the initial conditions $x = 0$ when $y = 0$ to fin $C$:

$- \frac{1}{4} \sqrt{9 - 0} = 0 + C \implies C = - \frac{3}{4}$

Hence, the Particular Solution is:

$- \frac{1}{4} \sqrt{9 - 4 {y}^{2}} = x - \frac{3}{4}$

And for an explicit solution:

$\sqrt{9 - 4 {y}^{2}} = 3 - 4 x$
$\therefore 9 - 4 {y}^{2} = {\left(3 - 4 x\right)}^{2}$ ..... $\left(\star\right)$
$\therefore 9 - 4 {y}^{2} = 9 - 24 x + 16 {x}^{2}$
$\therefore 4 {y}^{2} = 24 x - 16 {x}^{2}$
$\therefore {y}^{2} = 6 x - 4 {x}^{2}$
$\therefore y = \pm \sqrt{6 x - 4 {x}^{2}}$, confirming the given solution.

(NB, Care should be taken in interpreting the solution within the context of the model of the DE, as step $\left(\star\right)$ may introduce an invalid negative solution).

Sep 17, 2017

See below.

#### Explanation:

Making

$z = {\left(\frac{2}{3} y\right)}^{2}$ we get at the equivalent system

$\frac{9}{8} z ' - 3 \sqrt{1 - z} = 0$ or

$\frac{z '}{\sqrt{1 - z}} = 3 \times \frac{8}{9} = \frac{8}{3}$ or

$\frac{\mathrm{dz}}{\sqrt{1 - z}} = \frac{8}{3} \mathrm{dx}$ or after integrating

$- 2 \sqrt{1 - z} = \frac{8}{3} x + C$ and

${\left(\frac{4}{3} x + {C}_{1}\right)}^{2} = 1 - z \Rightarrow z = 1 - {\left(\frac{4}{3} x + {C}_{1}\right)}^{2} = {\left(\frac{2}{3} y\right)}^{2}$ and then

$y = \pm \frac{3}{2} \sqrt{1 - {\left(\frac{4}{3} x + {C}_{1}\right)}^{2}}$

Now with the initial conditions

$y \left(0\right) = \pm \frac{3}{2} \sqrt{1 - {C}_{1}^{2}} = 0 \Rightarrow {C}_{1} = \pm 1$ and

$y = \pm \frac{3}{2} \sqrt{1 - {\left(\frac{4}{3} x \pm 1\right)}^{2}}$