# How to solve the separable differential equation and find the particular solution satisfying the initial condition y(−4)=3 ?

## General Solution: $\textcolor{red}{{\left(4 y + 13\right)}^{\frac{1}{2}} - 2 x = {C}_{1}} \text{ }$
Particular Solution: $\textcolor{b l u e}{{\left(4 y + 13\right)}^{\frac{1}{2}} - 2 x = 13}$

#### Explanation:

From the given differential equation $y ' \left(x\right) = \sqrt{4 y \left(x\right) + 13}$

take note, that $y ' \left(x\right) = \frac{\mathrm{dy}}{\mathrm{dx}}$ and $y \left(x\right) = y$, therefore

$\frac{\mathrm{dy}}{\mathrm{dx}} = \sqrt{4 y + 13}$

divide both sides by $\sqrt{4 y + 13}$

$\frac{\mathrm{dy}}{\mathrm{dx}} \left(\frac{1}{\sqrt{4 y + 13}}\right) = \frac{\sqrt{4 y + 13}}{\sqrt{4 y + 13}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} \left(\frac{1}{\sqrt{4 y + 13}}\right) = 1$

Multiply both sides by $\mathrm{dx}$

$\mathrm{dx} \cdot \frac{\mathrm{dy}}{\mathrm{dx}} \left(\frac{1}{\sqrt{4 y + 13}}\right) = \mathrm{dx} \cdot 1$

$\cancel{\mathrm{dx}} \cdot \frac{\mathrm{dy}}{\cancel{\mathrm{dx}}} \left(\frac{1}{\sqrt{4 y + 13}}\right) = \mathrm{dx} \cdot 1$

$\frac{\mathrm{dy}}{\sqrt{4 y + 13}} = \mathrm{dx}$

transpose $\mathrm{dx}$ to the left side

$\frac{\mathrm{dy}}{\sqrt{4 y + 13}} - \mathrm{dx} = 0$

integrating on both sides we have the following results

$\int \frac{\mathrm{dy}}{\sqrt{4 y + 13}} - \int \mathrm{dx} = \int 0$

$\frac{1}{4} \cdot \int {\left(4 y + 13\right)}^{- \frac{1}{2}} \cdot 4 \cdot \mathrm{dy} - \int \mathrm{dx} = \int 0$

$\frac{1}{4} \cdot {\left(4 y + 13\right)}^{- \frac{1}{2} + 1} / \left(\left(1 - \frac{1}{2}\right)\right) - x = {C}_{0}$

$\frac{1}{2} \cdot {\left(4 y + 13\right)}^{\frac{1}{2}} - x = {C}_{0}$

${\left(4 y + 13\right)}^{\frac{1}{2}} - 2 x = 2 \cdot {C}_{0}$

$\textcolor{red}{{\left(4 y + 13\right)}^{\frac{1}{2}} - 2 x = {C}_{1}} \text{ }$General Solution

But $y \left(- 4\right) = 3$ means when $x = - 4$ , $y = 3$
We can now solve for ${C}_{1}$ to solve for the particular solution

${\left(4 y + 13\right)}^{\frac{1}{2}} - 2 x = {C}_{1}$

${\left(4 \left(3\right) + 13\right)}^{\frac{1}{2}} - 2 \left(- 4\right) = {C}_{1}$

${C}_{1} = 13$

Therefore , our particular solution is

$\textcolor{b l u e}{{\left(4 y + 13\right)}^{\frac{1}{2}} - 2 x = 13}$

God bless....I hope the explanation is useful.

May 14, 2016

$y = {x}^{2} + 13 x + 36$, with $y \ge - \frac{13}{4}$.

#### Explanation:

$y \ge - \frac{13}{4}$, to make $\sqrt{4 y + 13}$ real..

Rearranging,

$x ' \left(y\right) = \frac{1}{\sqrt{4 y + 13}}$

So, $x = \int \frac{1}{\sqrt{4 y + 13}} \mathrm{dy}$

$= \left(\frac{4}{2}\right) \sqrt{4 y + 13} + C$

Using y =3, when x = -4, C = -`13/2

So. $x = \left(\frac{1}{2}\right) \left(\sqrt{4 y + 13} - 13\right)$

Inversely. $y = \left(\frac{1}{4}\right) \left({\left(2 x + 13\right)}^{2} - 13\right) = {x}^{2} + 13 x + 36$