`dy/dx = (x^3-21x) / (5+4x-x^2)` Solve the differential equation

Monday, March 31, 2014

$\displaystyle\frac{{\left.{d}{y}}}{{\left.{d}{x}}}=\frac{{{{x}}^{{3}}-{21}{x}}}{{{5}+{4}{x}-{{x}}^{{2}}}}$ Solve the differential equation

To solve the differential equation of $\displaystyle\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}=\frac{{{{x}}^{{3}}-{2}{x}}}{{{5}+{4}{x}-{{x}}^{{2}}}}$ .we may express it in a form of variable separable differential equation: $\displaystyle{N}{\left({y}\right)}{\left.{d}{y}\right.}={M}{\left({x}\right)}{\left.{d}{x}\right.}$

$\displaystyle{\left.{d}{y}\right.}=\frac{{{{x}}^{{3}}-{2}{x}}}{{{5}+{4}{x}-{{x}}^{{2}}}}{\left.{d}{x}\right.}$

Then apply direct integration on both sides:

$\displaystyle\int{\left.{d}{y}\right.}=\int\frac{{{{x}}^{{3}}-{2}{x}}}{{{5}+{4}{x}-{{x}}^{{2}}}}{\left.{d}{x}\right.}$

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

For the right side, we may apply long division to expand:

$\displaystyle\int\frac{{{{x}}^{{3}}-{2}{x}}}{{{5}+{4}{x}-{{x}}^{{2}}}}{\left.{d}{x}\right.}=\int{\left[-{x}-{4}+\frac{{{20}}}{{{5}+{4}{x}-{{x}}^{{2}}}}\right]}{\left.{d}{x}\right.}$

Then apply basic integration property: $\displaystyle\int{\left({u}\pm{v}\right)}{\left.{d}{x}\right.}=\int{\left({u}\right)}{\left.{d}{x}\right.}\pm\int{\left({v}\right)}{\left.{d}{x}\right.}$

where we can integrate each term separately.

$\displaystyle\int{\left[-{x}-{4}+\frac{{{20}}}{{{5}+{4}{x}-{{x}}^{{2}}}}\right]}{\left.{d}{x}\right.}=\int{\left(-{x}\right)}{\left.{d}{x}\right.}-\int{4}{\left.{d}{x}\right.}+\int\frac{{{20}}}{{{5}+{4}{x}-{{x}}^{{2}}}}{\]}{\left.{d}{x}\right.}$

For the integration of $\displaystyle\int{\left(-{x}\right)}{\left.{d}{x}\right.}$ , we may apply Power Rule integration: $\displaystyle\int{{u}}^{{n}}{d}{u}=\frac{{{u}}^{{{n}+{1}}}}{{{n}+{1}}}+{C}$

$\displaystyle\int{\left(-{x}\right)}{\left.{d}{x}\right.}=-\int{x}{\left.{d}{x}\right.}$

$\displaystyle=-\frac{{{x}}^{{{1}+{1}}}}{{{1}+{1}}}$
$\displaystyle=-\frac{{{x}}^{{2}}}{{2}}$

For the integration of $\displaystyle-\int{4}{\left.{d}{x}\right.}$ , we may apply basic integration property: $\displaystyle\int{c}\cdot{f{{\left({x}\right)}}}{\left.{d}{x}\right.}={c}\int{f{{\left({x}\right)}}}{\left.{d}{x}\right.}$

$\displaystyle-\int{4}{\left.{d}{x}\right.}=-{4}\int{\left.{d}{x}\right.}$

$\displaystyle=-{4}{x}$

For the integration of $\displaystyle\int\frac{{20}}{{{5}+{4}{x}-{{x}}^{{2}}}}{\left.{d}{x}\right.}$ , we apply partial fractions:

$\displaystyle\frac{{{20}}}{{{5}+{4}{x}-{{x}}^{{2}}}}=\frac{{20}}{{{6}{\left({x}+{1}\right)}}}-\frac{{20}}{{{6}{\left({x}-{5}\right)}}}$

Then, $\displaystyle\int\frac{{{20}}}{{{5}+{4}{x}-{{x}}^{{2}}}}{\left.{d}{x}\right.}=\int{\left[\frac{{20}}{{{6}{\left({x}+{1}\right)}}}-\frac{{20}}{{{6}{\left({x}-{5}\right)}}}\right]}{\left.{d}{x}\right.}$

Apply basic integration property: $\displaystyle\int{\left({u}\pm{v}\right)}{\left.{d}{x}\right.}=\int{\left({u}\right)}{\left.{d}{x}\right.}\pm\int{\left({v}\right)}{\left.{d}{x}\right.}$

$\displaystyle\int{\left[\frac{{20}}{{{6}{\left({x}+{1}\right)}}}-\frac{{20}}{{{6}{\left({x}-{5}\right)}}}\right]}{\left.{d}{x}\right.}=\int\frac{{20}}{{{6}{\left({x}+{1}\right)}}}{\left.{d}{x}\right.}-\int\frac{{20}}{{{6}{\left({x}-{5}\right)}}}{\left.{d}{x}\right.}$

Apply apply the basic integration property: $\displaystyle\int{c}\cdot{f{{\left({x}\right)}}}{\left.{d}{x}\right.}={c}\int{f{{\left({x}\right)}}}{\left.{d}{x}\right.}$ and basic integration formula for logarithm: $\displaystyle\int\frac{{{d}{u}}}{{u}}={\ln}{\left|{u}\right|}+{C}$ .

$\displaystyle\int\frac{{20}}{{{6}{\left({x}+{1}\right)}}}{\left.{d}{x}\right.}-\int\frac{{20}}{{{6}{\left({x}-{5}\right)}}}{\left.{d}{x}\right.}={\left(\frac{{20}}{{6}}\right)}\int\frac{{1}}{{{x}+{1}}}{\left.{d}{x}\right.}-{\left(\frac{{20}}{{6}}\right)}\int\frac{{1}}{{{x}-{5}}}{\left.{d}{x}\right.}$

$\displaystyle={\left(\frac{{20}}{{6}}\right)}{\ln}{\left|{x}+{1}\right|}-{\left(\frac{{20}}{{6}}\right)}{\ln}{\left|{x}-{5}\right|}$

For the right side, we get:

$\displaystyle\int{\left[-{x}-{4}+\frac{{{20}}}{{{5}+{4}{x}-{{x}}^{{2}}}}\right]}{\left.{d}{x}\right.}=-\frac{{{x}}^{{2}}}{{2}}-{4}{x}+{\left(\frac{{20}}{{6}}\right)}{\ln}{\left|{x}+{1}\right|}-{\left(\frac{{20}}{{6}}\right)}{\ln}{\left|{x}-{5}\right|}+{C}$

Note: Just include the constant of integration "C" on one side as the arbitrary constant of a differential equation.

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

$\displaystyle{y}=-\frac{{{x}}^{{2}}}{{2}}-{4}{x}+{\left(\frac{{20}}{{6}}\right)}{\ln}{\left|{x}+{1}\right|}-{\left(\frac{{20}}{{6}}\right)}{\ln}{\left|{x}-{5}\right|}+{C}$

$\displaystyle{y}=-\frac{{{x}}^{{2}}}{{2}}-{4}{x}+{\left(\frac{{10}}{{3}}\right)}{\ln}{\left|{x}+{1}\right|}-{\left(\frac{{10}}{{3}}\right)}{\ln}{\left|{x}-{5}\right|}+{C}$

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Monday, March 31, 2014

$\displaystyle\frac{{\left.{d}{y}}}{{\left.{d}{x}}}=\frac{{{{x}}^{{3}}-{21}{x}}}{{{5}+{4}{x}-{{x}}^{{2}}}}$ Solve 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?

Monday, March 31, 2014

Solve the differential equation

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

$\displaystyle\frac{{\left.{d}{y}}}{{\left.{d}{x}}}=\frac{{{{x}}^{{3}}-{21}{x}}}{{{5}+{4}{x}-{{x}}^{{2}}}}$ Solve the differential equation

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