`dy/dx = (1-2x) / (4x-x^2)` Solve the differential equation

Monday, February 29, 2016

$\displaystyle\frac{{\left.{d}{y}}}{{\left.{d}{x}}}=\frac{{{1}-{2}{x}}}{{{4}{x}-{{x}}^{{2}}}}$ Solve the differential equation

$\displaystyle\frac{{\left.{d}{y}}}{{\left.{d}{x}}}=\frac{{{1}-{2}{x}}}{{{4}{x}-{{x}}^{{2}}}}$

This differential equation is separable since it can be written in the form

$\displaystyle{N}{\left({y}\right)}{\left.{d}{y}\right.}={M}{\left({x}\right)}{\left.{d}{x}\right.}$

Bringing together same variables on one side, the equation becomes

$\displaystyle{\left.{d}{y}\right.}=\frac{{{1}-{2}{x}}}{{{4}{x}-{{x}}^{{2}}}}{\left.{d}{x}\right.}$

Taking the integral of both sides, it turns into

$\displaystyle\int{\left.{d}{y}\right.}=\int\frac{{{1}-{2}{x}}}{{{4}{x}-{{x}}^{{2}}}}{\left.{d}{x}\right.}$

$\displaystyle{y}+{C}_{{1}}=\int\frac{{{1}-{2}{x}}}{{{4}{x}-{{x}}^{{2}}}}{\left.{d}{x}\right.}$

$\displaystyle{y}+{C}_{{1}}=\int\frac{{{1}-{2}{x}}}{{{x}{\left({4}-{x}\right)}}}{\left.{d}{x}\right.}$

To take the integral of right side, apply partial fraction decomposition.

$\displaystyle\frac{{{1}-{2}{x}}}{{{x}{\left({4}-{x}\right)}}}=\frac{{A}}{{x}}+\frac{{B}}{{{4}-{x}}}$

$\displaystyle{1}-{2}{x}={A}{\left({4}-{x}\right)}+{B}{x}$

Let x=0.

$\displaystyle{1}-{2}{\left({0}\right)}={A}{\left({4}-{0}\right)}+{B}{\left({0}\right)}$

$\displaystyle{1}={4}{A}$

$\displaystyle\frac{{1}}{{4}}={A}$

Let x=4.

$\displaystyle{1}-{2}{\left({4}\right)}={A}{\left({4}-{4}\right)}+{B}{\left({4}\right)}$

$\displaystyle-{7}={4}{B}$

$\displaystyle-\frac{{7}}{{4}}={B}$

$\displaystyle\frac{{1}}{{{2}{x}-{{x}}^{{2}}}}=\frac{{\frac{{1}}{{4}}}}{{x}}+\frac{{-\frac{{7}}{{4}}}}{{{4}-{x}}}$

$\displaystyle\frac{{1}}{{{2}{x}-{{x}}^{{2}}}}=\frac{{1}}{{{4}{x}}}+\frac{{-{7}}}{{{4}{\left({4}-{x}\right)}}}$

$\displaystyle\frac{{1}}{{{2}{x}-{{x}}^{{2}}}}=\frac{{1}}{{{4}{x}}}+\frac{{-{7}}}{{-{4}{\left({x}-{4}\right)}}}$

$\displaystyle\frac{{1}}{{{2}{x}-{{x}}^{{2}}}}=\frac{{1}}{{{4}{x}}}+\frac{{{7}}}{{{4}{\left({x}-{4}\right)}}}$

So the integrand at the right side decomposes to

$\displaystyle{y}+{C}_{{1}}=\int{\left(\frac{{1}}{{{4}{x}}}+\frac{{7}}{{{4}{\left({x}-{4}\right)}}}\right)}{\left.{d}{x}\right.}$

Then, apply the formula $\displaystyle\int\frac{{1}}{{u}}{d}{u}={\ln}{\left|{u}\right|}+{C}$ .

$\displaystyle{y}+{C}_{{1}}=\frac{{1}}{{4}}{\ln}{\left|{x}\right|}+\frac{{7}}{{4}}{\ln}{\left|{x}-{4}\right|}+{C}_{{2}}$

Isolating the y, the equation becomes

$\displaystyle{y}=\frac{{1}}{{4}}{\ln}{\left|{x}\right|}+\frac{{7}}{{4}}{\ln}{\left|{x}-{4}\right|}+{C}_{{2}}-{C}_{{1}}$

Since C1 and C2 represent any number, it can be expressed as a single constant C.

$\displaystyle{y}=\frac{{1}}{{4}}{\ln}{\left|{x}\right|}+\frac{{7}}{{4}}{\ln}{\left|{x}-{4}\right|}+{C}$

Therefore, the general solution of the given differential equation is $\displaystyle{y}=\frac{{1}}{{4}}{\ln}{\left|{x}\right|}+\frac{{7}}{{4}}{\ln}{\left|{x}-{4}\right|}+{C}$ .

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Monday, February 29, 2016

$\displaystyle\frac{{\left.{d}{y}}}{{\left.{d}{x}}}=\frac{{{1}-{2}{x}}}{{{4}{x}-{{x}}^{{2}}}}$ 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, February 29, 2016

Solve the differential equation

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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?

$\displaystyle\frac{{\left.{d}{y}}}{{\left.{d}{x}}}=\frac{{{1}-{2}{x}}}{{{4}{x}-{{x}}^{{2}}}}$ 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?