`dy/dx = 1/((x-1)sqrt(-4x^2+8x-1))` Solve the differential equation

Thursday, September 8, 2011

$\displaystyle\frac{{\left.{d}{y}}}{{\left.{d}{x}}}=\frac{{1}}{{{\left({x}-{1}\right)}\sqrt{{-{4}{{x}}^{{2}}+{8}{x}-{1}}}}}$ Solve the differential equation

The given problem $\displaystyle\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}=\frac{{1}}{{{\left({x}-{1}\right)}\sqrt{{-{4}{{x}}^{{2}}+{8}{x}+{1}}}}}$ is in form of a first order ordinary differential equation. To evaluate this, we may follow the variable separable differential equation: $\displaystyle{N}{\left({y}\right)}{\left.{d}{y}\right.}={M}{\left({x}\right)}{\left.{d}{x}\right.}$ .

$\displaystyle{\left.{d}{y}\right.}=\frac{{1}}{{{\left({x}-{1}\right)}\sqrt{{-{4}{{x}}^{{2}}+{8}{x}+{1}}}}}{\left.{d}{x}\right.}$

Apply direct integration on both sides:

$\displaystyle\int{\left.{d}{y}\right.}=\int\frac{{1}}{{{\left({x}-{1}\right)}\sqrt{{-{4}{{x}}^{{2}}+{8}{x}+{1}}}}}{\left.{d}{x}\right.}$

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

For the right side, we apply several substitutions to simplify it.

Let $\displaystyle{u}={\left({x}-{1}\right)}$ then $\displaystyle{x}={u}+{1}$ and $\displaystyle{d}{u}={\left.{d}{x}\right.}$ . The integral becomes:

$\displaystyle\int\frac{{1}}{{{\left({u}\right)}\sqrt{{-{4}{{x}}^{{2}}+{8}{x}+{1}}}}}{\left.{d}{x}\right.}=\int\frac{{1}}{{{u}\sqrt{{-{4}{{\left({u}+{1}\right)}}^{{2}}+{8}{\left({u}+{1}\right)}+{1}}}}}{d}{u}$

$\displaystyle=\int\frac{{1}}{{{u}\sqrt{{-{4}{\left({{u}}^{{2}}+{2}{u}+{1}\right)}+{8}{u}+{8}+{1}}}}}{d}{u}$

$\displaystyle=\int\frac{{1}}{{{u}\sqrt{{-{4}{{u}}^{{2}}-{8}{u}-{4}+{8}{u}+{8}+{1}}}}}{d}{u}$

$\displaystyle=\int\frac{{1}}{{{u}\sqrt{{-{4}{{u}}^{{2}}+{5}}}}}{d}{u}$

Let $\displaystyle{v}={{u}}^{{2}}$ then $\displaystyle{d}{v}={2}{u}{d}{u}$ or $\displaystyle\frac{{{d}{v}}}{{{2}{u}}}={d}{u}$ . The integral becomes:

$\displaystyle\int\frac{{1}}{{{u}\sqrt{{-{4}{{u}}^{{2}}+{5}}}}}{d}{u}=\int\frac{{1}}{{{u}\sqrt{{-{4}{v}+{5}}}}}\cdot\frac{{{d}{v}}}{{{2}{u}}}$

$\displaystyle=\int\frac{{{d}{v}}}{{{2}{{u}}^{{2}}\sqrt{{-{4}{v}+{5}}}}}$

$\displaystyle=\int\frac{{{d}{v}}}{{{2}{v}\sqrt{{-{4}{v}+{5}}}}}$

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

$\displaystyle\int\frac{{{d}{v}}}{{{2}{v}\sqrt{{-{4}{v}+{5}}}}}={\left(\frac{{1}}{{2}}\right)}\int\frac{{{d}{v}}}{{{v}\sqrt{{-{4}{v}+{5}}}}}$

Let $\displaystyle{w}=\sqrt{{-{4}{v}+{5}}}$ then $\displaystyle{v}=\frac{{{5}-{{w}}^{{2}}}}{{4}}$ and $\displaystyle{d}{w}=-\frac{{2}}{\sqrt{{-{4}{v}+{5}}}}{d}{v}$ or

$\displaystyle\frac{{{d}{w}}}{{-{2}}}=\frac{{1}}{\sqrt{{-{4}{v}+{5}}}}{d}{v}$

The integral becomes:

$\displaystyle{\left(\frac{{1}}{{2}}\right)}\int\frac{{{d}{v}}}{{{v}\sqrt{{-{4}{v}+{5}}}}}={\left(\frac{{1}}{{2}}\right)}\int\frac{{1}}{{v}}\cdot\frac{{{d}{v}}}{\sqrt{{-{4}{v}+{5}}}}$

$\displaystyle={\left(\frac{{1}}{{2}}\right)}\int\frac{{1}}{{\frac{{{5}-{{w}}^{{2}}}}{{4}}}}\cdot\frac{{{d}{w}}}{{-{2}}}$

$\displaystyle={\left(\frac{{1}}{{2}}\right)}\int{1}\cdot\frac{{4}}{{{5}-{{w}}^{{2}}}}\cdot\frac{{{d}{w}}}{{-{2}}}$

$\displaystyle={\left(\frac{{1}}{{2}}\right)}\int-\frac{{2}}{{{5}-{{w}}^{{2}}}}{d}{w}$

$\displaystyle={\left(\frac{{1}}{{2}}\right)}\cdot-{2}\int\frac{{1}}{{{5}-{{w}}^{{2}}}}{d}{w}$

$\displaystyle={\left(-{1}\right)}\int\frac{{1}}{{{5}-{{w}}^{{2}}}}{d}{w}$

Apply basic integration formula for inverse hyperbolic tangent function:

$\displaystyle\int\frac{{{d}{u}}}{{{{a}}^{{2}}-{{u}}^{{2}}}}={\left(\frac{{1}}{{a}}\right)}{\arctan{{h}}}{\left(\frac{{u}}{{a}}\right)}+{C}$

Then, with corresponding values as: $\displaystyle{{a}}^{{2}}={5}$ and $\displaystyle{{u}}^{{2}}={{u}}^{{2}}$ , we get: $\displaystyle{a}=\sqrt{{{5}}}$ and $\displaystyle{u}={w}$

$\displaystyle{\left(-{1}\right)}\int\frac{{1}}{{{5}-{{w}}^{{2}}}}{d}{w}=-\frac{{1}}{\sqrt{{{5}}}}{\arctan{{h}}}{\left(\frac{{w}}{\sqrt{{{5}}}}\right)}+{C}$

Recall $\displaystyle{w}=\sqrt{{-{4}{v}+{5}}}$ and $\displaystyle{v}={{u}}^{{2}}$ then $\displaystyle{w}=\sqrt{{-{4}{{u}}^{{2}}+{5}}}.$

Plug-in $\displaystyle{u}={\left({x}-{1}\right)}$ on $\displaystyle{w}=\sqrt{{-{4}{{u}}^{{2}}+{5}}}$ , we get:

$\displaystyle{w}=\sqrt{{-{4}{{\left({x}-{1}\right)}}^{{2}}+{5}}}$

$\displaystyle{w}=\sqrt{{-{4}{\left({{x}}^{{2}}-{2}{x}+{1}\right)}+{5}}}$

$\displaystyle{w}=\sqrt{{-{4}{{x}}^{{2}}+{8}{x}-{4}+{5}}}$

$\displaystyle{w}=\sqrt{{-{4}{{x}}^{{2}}+{8}{x}+{1}}}$

Plug-in $\displaystyle{w}=\sqrt{{-{4}{{x}}^{{2}}+{8}{x}+{1}}}$ on $\displaystyle-\frac{{1}}{\sqrt{{{5}}}}{\arctan{{h}}}{\left(\frac{{w}}{\sqrt{{{5}}}}\right)}+{C}$ , we get:

$\displaystyle\int\frac{{1}}{{{\left({x}-{1}\right)}\sqrt{{-{4}{{x}}^{{2}}+{8}{x}+{1}}}}}{\left.{d}{x}\right.}=\frac{{1}}{\sqrt{{{5}}}}{\arctan{{h}}}{\left(\frac{\sqrt{{-{4}{{x}}^{{2}}+{8}{x}+{1}}}}{\sqrt{{{5}}}}\right)}+{C}$

$\displaystyle=-\frac{{1}}{\sqrt{{{5}}}}{\arctan{{h}}}{\left(\frac{\sqrt{{-{4}{{x}}^{{2}}+{8}{x}+{1}}}}{{5}}\right)}+{C}$

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

$\displaystyle{y}=-\frac{{1}}{\sqrt{{{5}}}}{\arctan{{h}}}{\left(\frac{\sqrt{{-{4}{{x}}^{{2}}+{8}{x}+{1}}}}{{5}}\right)}+{C}$

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Thursday, September 8, 2011

$\displaystyle\frac{{\left.{d}{y}}}{{\left.{d}{x}}}=\frac{{1}}{{{\left({x}-{1}\right)}\sqrt{{-{4}{{x}}^{{2}}+{8}{x}-{1}}}}}$ 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?

Thursday, September 8, 2011

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?

$\displaystyle\frac{{\left.{d}{y}}}{{\left.{d}{x}}}=\frac{{1}}{{{\left({x}-{1}\right)}\sqrt{{-{4}{{x}}^{{2}}+{8}{x}-{1}}}}}$ 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?