# Determine the solution to the initial value differential equation y ? + 9 y 2 e 2 x = 0...

## Question:

Determine the solution to the initial value differential equation

{eq}y' \ + \ 9y^{2}e^{2x} \ = \ 0, \ y\left ( 0 \right ) \ = \ 1 {/eq}

y(x) = _____

## First order variable separable ordinary differential equation:

A separable differential equation is any differential equation that we can write in the following form.

{eq}\frac{{dy}}{{dx}} = F(y)G(x) {/eq}

we can re-write above equation as,

{eq}\frac{{dy}}{{F(y)}} = G(x)dx {/eq}

On integrating both side,

{eq}\int {\frac{{dy}}{{F(y)}}} = \int {G(x)} dx+c. {/eq}

where {eq}c {/eq} is the constant of integration.

We use the following formula of integration:

{eq}\begin{align} \int {e^{ax}} dx &= \frac{e^{ax}}{a} + c \hfill \\ \int {{x^n}} dx &= \frac{{{x^{n + 1}}}}{{n + 1}} + c, \hfill \\ \end{align} {/eq}

where {eq}c {/eq} is the constant of integration.

Given D.E. {eq}y' \ + \ 9y^{2}e^{2x} \ = \ 0, {/eq}

we can re-write above differential equation as,

{eq}\eqalign{ & \frac{{dy}}{{dx}} = - 9{y^2}{e^{2x}} \cr & \frac{{dy}}{{{y^2}}} = - 9{e^{2x}}dx, \cr} {/eq}

On integrating both sides,

{eq}\eqalign{ & \int {\frac{{dy}}{{{y^2}}} = - 9\int {{e^{2x}}dx} } \cr & - \frac{1}{y} = - 9\left( {\frac{{{e^{2x}}}}{2}} \right) + c \cr & - \frac{1}{y} = - \frac{9}{2}{e^{2x}} + c, \cr} {/eq}

Using initial condition {eq}y\left ( 0 \right ) \ = \ 1 {/eq}

{eq}\eqalign{ & - 1 = - \frac{9}{2}{e^0} + c \cr & c = \frac{7}{2}, \cr} {/eq}

Hence,

{eq}\eqalign{ & - \frac{1}{y} = - \frac{9}{2}{e^{2x}} + \frac{7}{2} \cr & y(x) = \frac{2}{{9{e^{2x}} - 7}}. \cr} {/eq}