# How to solve the separable differential equation and to the initial condition: y(0)=1 ?

May 6, 2016

Get the $x$s on one side and $y$s on the other, integrate, and simplify to get $y = \sqrt{\frac{5}{2} \left(\sqrt{{x}^{2} + 1}\right) - \frac{3}{2}}$.

#### Explanation:

We will begin to solve this first-order separable differential equation by separating it (no surprise there).

If we add $8 y \sqrt{{x}^{2} + 1} \frac{\mathrm{dy}}{\mathrm{dx}}$ to both sides we get:
$10 x = 8 y \sqrt{{x}^{2} + 1} \frac{\mathrm{dy}}{\mathrm{dx}}$

Now divide by $8 \sqrt{{x}^{2} + 1}$ to get:
$\frac{10 x}{8 \sqrt{{x}^{2} + 1}} = y \frac{\mathrm{dy}}{\mathrm{dx}}$

Multiply by $\mathrm{dx}$ to finally end up with:
$\frac{10 x}{8 \sqrt{{x}^{2} + 1}} \mathrm{dx} = y \mathrm{dy}$

Yay! We've separated the equation: we have $x$ on one side and $y$ on the other. The only thing that's left is to integrate both sides:
$\int \frac{10 x}{8 \sqrt{{x}^{2} + 1}} \mathrm{dx} = \int y \mathrm{dy}$

$\int \frac{10 x}{8 \sqrt{{x}^{2} + 1}} \mathrm{dx}$

First, take out a $\frac{10}{8}$ and simplify:
$\frac{5}{4} \int \frac{x}{\sqrt{{x}^{2} + 1}} \mathrm{dx}$

Now we can apply the substitution $u = {x}^{2} + 1 \to \frac{\mathrm{du}}{\mathrm{dx}} = 2 x \to \mathrm{du} = 2 x \mathrm{dx}$

In order to apply the substitution, we need to multiply the inside of the integral by $2$, so we end up with $2 x$. If we do that, we need to multiply the outside of the integral by $\frac{1}{2}$:
$\frac{5}{8} \int \frac{2 x}{\sqrt{{x}^{2} + 1}} \mathrm{dx}$

We can go ahead and substitute now:
$\frac{5}{8} \int \frac{\mathrm{du}}{\sqrt{u}}$

Noticing that this is equivalent to $\frac{5}{8} \int {u}^{- \frac{1}{2}} \mathrm{du}$ and using the reverse power rule, we end up with a solution of:
$2 \cdot \frac{5}{8} \left({u}^{\frac{1}{2}}\right) + C = \frac{5}{4} \sqrt{u} + C$

Since $u = {x}^{2} + 1$, we back-substitute to get our final answer:
$\frac{5}{4} \sqrt{{x}^{2} + 1} + C$

As for the other integral, $\int y \mathrm{dy}$, well, that's just ${y}^{2} / 2 + C$.

Alright, we've solved our integrals so we now have:
$\frac{5}{4} \sqrt{{x}^{2} + 1} + C = {y}^{2} / 2 + C$

Doing a little algebra to solve for $y$ yields:
$y = \sqrt{\frac{5}{2} \left(\sqrt{{x}^{2} + 1}\right) + C}$

Note: Remember that manipulating the integration constant $C$ makes no difference. Whatever we do to it, it's still a constant.

Now we apply the initial condition $y \left(0\right) = 1$ to solve for $C$:
$y = \sqrt{\frac{5}{2} \left(\sqrt{{x}^{2} + 1}\right) + C}$
$1 = \sqrt{\frac{5}{2} \left(\sqrt{{\left(0\right)}^{2} + 1}\right) + C}$
$1 = \frac{5}{2} + C$
$C = - \frac{3}{2}$

Therefore our solution is $y = \sqrt{\frac{5}{2} \left(\sqrt{{x}^{2} + 1}\right) - \frac{3}{2}}$.