`yy' - 2e^x = 0 , y(0) = 3` Find the particular solution that satisfies the initial condition

Sunday, December 27, 2009

$\displaystyle{y}{y}'-{2}{{e}}^{{x}}={0},{y}{\left({0}\right)}={3}$ Find the particular solution that satisfies the initial condition

For the given problem: $\displaystyle{y}{y}'-{2}{{e}}^{{x}}={0}$ , we can evaluate this by applying variable separable differential equation in which we express it in a form of $\displaystyle{f{{\left({y}\right)}}}{\left.{d}{y}\right.}={f{{\left({x}\right)}}}{\left.{d}{x}\right.}$ .

Then, $\displaystyle{y}{y}'-{2}{{e}}^{{x}}={0}$ can be rearrange into $\displaystyle{y}{y}'={2}{{e}}^{{x}}$

Express y' as (dy)/(dx):

$\displaystyle{y}\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}={2}{{e}}^{{x}}$

Apply direct integration in the form of $\displaystyle\int{f{{\left({y}\right)}}}{\left.{d}{y}\right.}=\int{f{{\left({x}\right)}}}{\left.{d}{x}\right.}$ :

$\displaystyle{y}\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}={2}{{e}}^{{x}}$

$\displaystyle{y}{\left.{d}{y}\right.}={2}{{e}}^{{x}}{\left.{d}{x}\right.}$

$\displaystyle\int{y}{\left.{d}{y}\right.}=\int{2}{{e}}^{{x}}{\left.{d}{x}\right.}$

For the left side, we apply Power Rule integration: $\displaystyle\int{{u}}^{{n}}{d}{u}=\frac{{{u}}^{{{n}+{1}}}}{{{n}+{1}}}$ .

$\displaystyle\int{y}{\left.{d}{y}\right.}=\frac{{{y}}^{{{1}+{1}}}}{{{1}+{1}}}$

$\displaystyle=\frac{{{y}}^{{2}}}{{2}}$

For the right side, we apply basic integration property: $\displaystyle\int{c}\cdot{f{{\left({x}\right)}}}{\left.{d}{x}\right.}={c}\int{f{{\left({x}\right)}}}{\left.{d}{x}\right.}$ and basic integration formula for exponential function: $\displaystyle\int{{e}}^{{u}}{d}{u}={{e}}^{{u}}+{C}$ on the right side.

$\displaystyle\int{2}{{e}}^{{x}}{\left.{d}{x}\right.}={2}\int{{e}}^{{x}}{\left.{d}{x}\right.}$

$\displaystyle={2}{{e}}^{{x}}+{C}$

Combining the results for the general solution of differential equation:

$\displaystyle\frac{{{y}}^{{2}}}{{2}}={2}{{e}}^{{x}}+{C}$

$\displaystyle{2}\cdot{\left[\frac{{{y}}^{{2}}}{{2}}\right]}={2}\cdot{\left[{2}{{e}}^{{x}}\right]}+{2}\cdot{C}$

Let $\displaystyle{2}\cdot{C}={C}$ . Just a constant.

$\displaystyle{{y}}^{{2}}={4}{{e}}^{{x}}+{C}$

To find the particular solution we consider the initial condition $\displaystyle{y}{\left({0}\right)}={3}$ which implies $\displaystyle{x}={0}$ and $\displaystyle{y}={3}$ .