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

Tuesday, April 8, 2014

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

An ordinary differential equation (ODE) is differential equation for the derivative of a function of one variable. When an ODE is in a form of $\displaystyle{y}'={f{{\left({x},{y}\right)}}}$ , this is just a first order ordinary differential equation.

We may express $\displaystyle{y}'$ as $\displaystyle\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}$ to write in a form of $\displaystyle\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}={f{{\left({x},{y}\right)}}}$ and apply variable separable differential equation: $\displaystyle{N}{\left({y}\right)}{\left.{d}{y}\right.}={M}{\left({x}\right)}{\left.{d}{x}\right.}$ .

The given problem: $\displaystyle\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}={x}{{e}}^{{{{x}}^{{2}}}}$ can be rearrange as:

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

Apply direct integration on both sides:

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

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: $\displaystyle{u}={{x}}^{{2}}$ then $\displaystyle{d}{u}={2}{x}{\left.{d}{x}\right.}$ or $\displaystyle\frac{{{d}{u}}}{{2}}={x}{\left.{d}{x}\right.}$ .

The integral becomes:

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

$\displaystyle=\int{{e}}^{{{u}}}\cdot\frac{{{d}{u}}}{{2}}$

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

$\displaystyle\int{{e}}^{{{u}}}\cdot\frac{{{d}{u}}}{{2}}={\left(\frac{{1}}{{2}}\right)}\int{{e}}^{{{u}}}{d}{u}$

Apply basic integration formula for exponential function:

$\displaystyle{\left(\frac{{1}}{{2}}\right)}\int{{e}}^{{{u}}}{d}{u}={\left(\frac{{1}}{{2}}\right)}{{e}}^{{{u}}}+{C}$

Plug-in $\displaystyle{u}={{x}}^{{2}}$ on $\displaystyle{\left(\frac{{1}}{{2}}\right)}{{e}}^{{{u}}}+{C}$ , we get:

$\displaystyle\int{x}{{e}}^{{{{x}}^{{2}}}}{\left.{d}{x}\right.}={\left(\frac{{1}}{{2}}\right)}{{e}}^{{{{x}}^{{2}}}}+{C}$

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

$\displaystyle{y}={\left(\frac{{1}}{{2}}\right)}{{e}}^{{{{x}}^{{2}}}}+{C}$

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

Blog

Tuesday, April 8, 2014

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

No comments:

Post a Comment

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

Tuesday, April 8, 2014

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?

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