`dy/dx = 5e^(-x/2)` Use integration to find a general solution to the differential equation

Friday, April 17, 2015

$\displaystyle\frac{{\left.{d}{y}}}{{\left.{d}{x}}}={5}{{e}}^{{-\frac{{x}}{{2}}}}$ Use integration to find a general solution to the differential equation

The given problem: $\displaystyle\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}={5}{{e}}^{{-\frac{{x}}{{2}}}}$ is in form of a first order ordinary differential equation. To evaluate this, we may follow the variable separable differential equation: $\displaystyle{N}{\left({y}\right)}{\left.{d}{y}\right.}={M}{\left({x}\right)}{\left.{d}{x}\right.}$

Cross-multiply $\displaystyle{\left.{d}{x}\right.}$ to the other side, we get:

$\displaystyle{\left.{d}{y}\right.}={5}{{e}}^{{-\frac{{x}}{{2}}}}{\left.{d}{x}\right.}$

In this form, we may now proceed to direct integration on both sides:

$\displaystyle\int{\left.{d}{y}\right.}=\int{5}{{e}}^{{-\frac{{x}}{{2}}}}{\left.{d}{x}\right.}$

For the left side, we apply basic integration property: $\displaystyle\int{\left({\left.{d}{y}\right.}\right)}={y}$ .

For the right side, we may apply u-substitution by letting: $\displaystyle{u}=-\frac{{x}}{{2}}$ then $\displaystyle{d}{u}=-\frac{{1}}{{2}}{\left.{d}{x}\right.}$ or $\displaystyle-{2}{d}{u}={\left.{d}{x}\right.}$ .

Plug-in the values: $\displaystyle-\frac{{x}}{{2}}={u}$ and $\displaystyle{\left.{d}{x}\right.}=-{2}{d}{u}$ , we get:

$\displaystyle\int{5}{{e}}^{{-\frac{{x}}{{2}}}}{\left.{d}{x}\right.}=\int{5}{{e}}^{{{u}}}\cdot{\left(-{2}{d}{u}\right)}$

$\displaystyle=\int-{10}{{e}}^{{{u}}}{d}{u}$

Apply the basic integration property: $\displaystyle\int{c}\cdot{f{{\left({x}\right)}}}{\left.{d}{x}\right.}={c}\int{f{{\left({x}\right)}}}{\left.{d}{x}\right.}$ .

$\displaystyle\int-{10}{{e}}^{{{u}}}{d}{u}={\left(-{10}\right)}\int{{e}}^{{{u}}}{d}{u}$

Apply basic integration formula for exponential function:

$\displaystyle{\left(-{10}\right)}\int{{e}}^{{{u}}}{d}{u}=-{10}{{e}}^{{{u}}}+{C}$

Plug-in $\displaystyle{u}=-\frac{{x}}{{2}}$ on $\displaystyle-{10}{{e}}^{{{u}}}+{C}$ , we get:

$\displaystyle\int{5}{{e}}^{{-\frac{{x}}{{2}}}}{\left.{d}{x}\right.}=-{10}{{e}}^{{-\frac{{x}}{{2}}}}+{C}$

Combining the results from both sides, we get the general solution of differential equation as:

$\displaystyle{y}=-{10}{{e}}^{{-\frac{{x}}{{2}}}}+{C}$

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Friday, April 17, 2015

$\displaystyle\frac{{\left.{d}{y}}}{{\left.{d}{x}}}={5}{{e}}^{{-\frac{{x}}{{2}}}}$ Use integration to find a general solution to the differential equation

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Thomas Jefferson's election in 1800 is sometimes called the Revolution of 1800. Why could it be described in this way?

Friday, April 17, 2015

Use integration to find a general solution to the differential equation

No comments:

Post a Comment

Thomas Jefferson&#39;s election in 1800 is sometimes called the Revolution of 1800. Why could it be described in this way?

$\displaystyle\frac{{\left.{d}{y}}}{{\left.{d}{x}}}={5}{{e}}^{{-\frac{{x}}{{2}}}}$ Use integration to find a general solution to the differential equation

Thomas Jefferson's election in 1800 is sometimes called the Revolution of 1800. Why could it be described in this way?