`y' = -sqrt(x)/(4y)` Solve the differential equation.

Tuesday, December 16, 2014

$\displaystyle{y}'=-\frac{\sqrt{{{x}}}}{{{4}{y}}}$ Solve the differential equation.

An ordinary differential equation (ODE) is differential equation for the derivative of a function of one variable. When an ODE is in a form of $\displaystyle{y}'={f{{\left({x},{y}\right)}}}$ , this is just a first order ordinary differential equation.

The given problem: $\displaystyle{y}'=-\frac{\sqrt{{{x}}}}{{{4}{y}}}$ is in a form of $\displaystyle{y}'={f{{\left({x},{y}\right)}}}$ .

To evaluate this, we may express $\displaystyle{y}'$ as $\displaystyle\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}$ .

The problem becomes:

$\displaystyle\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}=-\frac{\sqrt{{{x}}}}{{{4}{y}}}$

We may apply the variable separable differential equation: $\displaystyle{N}{\left({y}\right)}{\left.{d}{y}\right.}={M}{\left({x}\right)}{d}$ x.

Cross-multiply $\displaystyle{\left.{d}{x}\right.}$ to the right side:

$\displaystyle{\left.{d}{y}\right.}=-\frac{\sqrt{{{x}}}}{{{4}{y}}}{\left.{d}{x}\right.}$

Cross-multiply $\displaystyle{4}{y}$ to the left side:

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

Apply direct integration on both sides:

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

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

$\displaystyle{4}\int{y}{\left.{d}{y}\right.}={\left(-{1}\right)}\int\sqrt{{{x}}}{\left.{d}{x}\right.}$

For the left side, we apply the Power Rule for integration: $\displaystyle\int{{u}}^{{n}}{d}{u}=\frac{{{u}}^{{{n}+{1}}}}{{{n}+{1}}}+{C}$ .

$\displaystyle{4}\int{y}{\left.{d}{y}\right.}={4}\cdot\frac{{{y}}^{{{1}+{1}}}}{{{1}+{1}}}$

$\displaystyle={4}\cdot\frac{{{y}}^{{2}}}{{2}}$

$\displaystyle={2}{{y}}^{{2}}$

For the right side, we apply Law of Exponent: $\displaystyle\sqrt{{{x}}}={{x}}^{{\frac{{1}}{{2}}}}$ before applying the Power Rule for integration: $\displaystyle\int{{u}}^{{n}}{d}{u}=\frac{{{u}}^{{{n}+{1}}}}{{{n}+{1}}}+{C}$ .

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

$\displaystyle=-\frac{{{x}}^{{\frac{{3}}{{2}}}}}{{\frac{{3}}{{2}}}}+{C}$

$\displaystyle=-{{x}}^{{\frac{{3}}{{2}}}}\cdot{\left(\frac{{2}}{{3}}\right)}+{C}$

$\displaystyle=-\frac{{{2}{{x}}^{{\frac{{3}}{{2}}}}}}{{3}}+{C}$

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

$\displaystyle{2}{{y}}^{{2}}=-\frac{{{2}{{x}}^{{\frac{{3}}{{2}}}}}}{{3}}+{C}$

We may simplify it as:

$\displaystyle{\left(\frac{{1}}{{2}}\right)}{\left[{2}{{y}}^{{2}}\right]}={\left(\frac{{1}}{{2}}\right)}{\left[-\frac{{{2}{{x}}^{{\frac{{3}}{{2}}}}}}{{3}}+{C}\right]}$

$\displaystyle{{y}}^{{2}}=-\frac{{{2}{{x}}^{{\frac{{3}}{{2}}}}}}{{6}}+\frac{{C}}{{2}}$

$\displaystyle{y}=\pm\sqrt{{-\frac{{{2}{{x}}^{{\frac{{3}}{{2}}}}}}{{6}}+\frac{{C}}{{2}}}}$

$\displaystyle{y}=\pm\sqrt{{-\frac{{{{x}}^{{\frac{{3}}{{2}}}}}}{{3}}+\frac{{C}}{{2}}}}$

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Tuesday, December 16, 2014

$\displaystyle{y}'=-\frac{\sqrt{{{x}}}}{{{4}{y}}}$ 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?

Tuesday, December 16, 2014

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?

$\displaystyle{y}'=-\frac{\sqrt{{{x}}}}{{{4}{y}}}$ 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?