`sqrt(1-4x^2)y' = x` Find the general solution of the differential equation

Tuesday, September 20, 2016

$\displaystyle\sqrt{{{1}-{4}{{x}}^{{2}}}}{y}'={x}$ Find the general solution of the differential equation

To be able to evaluate the problem: $\displaystyle\sqrt{{{1}-{4}{{x}}^{{2}}}}{y}'={x}$ , we express in a form of $\displaystyle{y}'={f{{\left({x}\right)}}}$ .

To do this, we divide both sides by $\displaystyle\sqrt{{{1}-{4}{{x}}^{{2}}}}$ .

$\displaystyle{y}'=\frac{{x}}{\sqrt{{{1}-{4}{{x}}^{{2}}}}}$

The general solution of a differential equation in a form of $\displaystyle{y}'={f{{\left({x}\right)}}}$ can

be evaluated using direct integration. We can denote y' as $\displaystyle\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}$ .

Then,

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

This is the same as $\displaystyle{\left({\left.{d}{y}\right.}\right)}=\frac{{x}}{\sqrt{{{1}-{4}{{x}}^{{2}}}}}{\left.{d}{x}\right.}$

Apply direct integration on both sides:

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

For the right side, we apply u-substitution using $\displaystyle{u}={1}-{4}{{x}}^{{2}}$ then $\displaystyle{d}{u}=-{8}{x}{\left.{d}{x}\right.}$ or $\displaystyle\frac{{{d}{u}}}{{-{8}}}={x}{\left.{d}{x}\right.}$ .

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

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

$\displaystyle\int\frac{{1}}{\sqrt{{{u}}}}\cdot\frac{{{d}{u}}}{{-{8}}}=-\frac{{1}}{{8}}\int\frac{{1}}{\sqrt{{{u}}}}{d}{u}$

Applying Law of Exponents: $\displaystyle\sqrt{{{x}}}=\frac{{{x}}^{{1}}}{{2}}$ and $\displaystyle\frac{{1}}{{{x}}^{{n}}}={{x}}^{{-{{n}}}}$ :

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

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

Applying the Power Rule for integration: $\displaystyle\int{{x}}^{{n}}=\frac{{{x}}^{{{n}+{1}}}}{{{n}+{1}}}+{C}$ .

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

$\displaystyle=-\frac{{1}}{{8}}\frac{{{u}}^{{\frac{{1}}{{2}}}}}{{\frac{{1}}{{2}}}}+{C}$

$\displaystyle=-\frac{{1}}{{8}}{{u}}^{{\frac{{1}}{{2}}}}\cdot{\left(\frac{{2}}{{1}}\right)}+{C}$

$\displaystyle=-\frac{{2}}{{8}}{{u}}^{{\frac{{1}}{{2}}}}+{C}$

$\displaystyle=-\frac{{1}}{{4}}{{u}}^{{\frac{{1}}{{2}}}}+{C}{\quad\text{or}\quad}-\frac{{1}}{{4}}\sqrt{{{u}}}+{C}$

Plug-in $\displaystyle{u}={1}-{4}{{x}}^{{2}}$ in $\displaystyle-\frac{{1}}{{4}}{{u}}^{{\frac{{1}}{{2}}}}$ , we get:

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

Combining the results, we get the general solution for differential equation

$\displaystyle{\left(\sqrt{{{1}-{4}{{x}}^{{2}}}}{y}'={x}\right)}$

as:

$\displaystyle{y}=-\frac{{1}}{{4}}\sqrt{{{1}-{4}{{x}}^{{2}}}}+{C}$

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Tuesday, September 20, 2016

$\displaystyle\sqrt{{{1}-{4}{{x}}^{{2}}}}{y}'={x}$ Find the general solution of 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?

Tuesday, September 20, 2016

Find the general solution of 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?

$\displaystyle\sqrt{{{1}-{4}{{x}}^{{2}}}}{y}'={x}$ Find the general solution of the differential equation

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