`dy/dx = 2xsqrt(4x^2+1)` Use integration to find a general solution to the differential equation

Tuesday, March 18, 2014

$\displaystyle\frac{{\left.{d}{y}}}{{\left.{d}{x}}}={2}{x}\sqrt{{{4}{{x}}^{{2}}+{1}}}$ Use integration to find a general solution to the differential equation

For the given problem: $\displaystyle\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}={2}{x}\sqrt{{{4}{{x}}^{{2}}+{1}}}$ is a first order ordinary differential equation in a form of $\displaystyle\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}={f{{\left({x},{y}\right)}}}$ .

To evaluate this, we rearrange it in a form of variable separable differential equation: $\displaystyle{N}{\left({y}\right)}{\left.{d}{y}\right.}={M}{\left({x}\right)}{\left.{d}{x}\right.}$ .

Cross-multiply $\displaystyle{\left.{d}{x}\right.}$ to the right side: $\displaystyle{\left.{d}{y}\right.}={2}{x}\sqrt{{{4}{{x}}^{{2}}+{1}}}{\left.{d}{x}\right.}$ .

Apply direct integration on both sides: $\displaystyle\int{\left.{d}{y}\right.}=\int{2}{x}\sqrt{{{4}{{x}}^{{2}}+{1}}}{\left.{d}{x}\right.}$ .

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

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

The integral becomes:

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

$\displaystyle=\int\frac{{\sqrt{{{u}}}{d}{u}}}{{4}}$

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

$\displaystyle\int\frac{{\sqrt{{{u}}}{d}{u}}}{{4}}=\frac{{1}}{{4}}\int\sqrt{{{u}}}{d}{u}$

Apply Law of Exponent: $\displaystyle\sqrt{{{x}}}={{x}}^{{\frac{{1}}{{2}}}}$ and Power Rule for integration : int $\displaystyle{{x}}^{{n}}=\frac{{{x}}^{{{n}+{1}}}}{{{n}+{1}}}+{C}$ .

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

$\displaystyle={\left(\frac{{1}}{{4}}\right)}\frac{{{u}}^{{\frac{{1}}{{2}}+{1}}}}{{\frac{{1}}{{2}}+{1}}}+{C}$

$\displaystyle={\left(\frac{{1}}{{4}}\right)}\frac{{{u}}^{{\frac{{3}}{{2}}}}}{{{\left(\frac{{3}}{{2}}\right)}}}+{C}$

$\displaystyle={\left(\frac{{1}}{{4}}\right)}{{u}}^{{\frac{{3}}{{2}}}}\cdot{\left(\frac{{2}}{{3}}\right)}+{C}$

$\displaystyle=\frac{{{u}}^{{\frac{{3}}{{2}}}}}{{6}}+{C}$

Plug-in $\displaystyle{u}={4}{{x}}^{{2}}+{1}$ on $\displaystyle\frac{{{u}}^{{\frac{{3}}{{2}}}}}{{6}}+{C}$ , we get:

$\displaystyle\int{2}{x}\sqrt{{{4}{{x}}^{{2}}+{1}}}{\left.{d}{x}\right.}=\frac{{{\left({4}{{x}}^{{2}}+{1}\right)}}^{{\frac{{3}}{{2}}}}}{{6}}+{C}$

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

$\displaystyle{y}=\frac{{{\left({4}{{x}}^{{2}}+{1}\right)}}^{{\frac{{3}}{{2}}}}}{{6}}+{C}$

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Tuesday, March 18, 2014

$\displaystyle\frac{{\left.{d}{y}}}{{\left.{d}{x}}}={2}{x}\sqrt{{{4}{{x}}^{{2}}+{1}}}$ Use integration to find a general solution to 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, March 18, 2014

Use integration to find a general solution to 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?

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