`(8, 2) , y' = 2y/(3x)` Find an equation of the graph that passes through the point and has the given slope

Friday, March 5, 2010

$\displaystyle{\left({8},{2}\right)},{y}'={2}\frac{{y}}{{{3}{x}}}$ Find an equation of the graph that passes through the point and has the given slope

The given slope equation: $\displaystyle{y}'={2}\frac{{y}}{{{3}{x}}}$ is in form of first order ordinary differential equation. In order to evaluate this, we let $\displaystyle{y}'$ as $\displaystyle\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}$ .

$\displaystyle\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}={2}\frac{{y}}{{{3}{x}}}$

Then, express as a variable separable differential equation: $\displaystyle{N}{\left({y}\right)}{\left.{d}{y}\right.}={M}{\left({x}\right)}{\left.{d}{x}\right.}$ .

To accomplish this, we cross-multiply $\displaystyle{\left.{d}{x}\right.}$ to the other side.

$\displaystyle{\left.{d}{y}\right.}={2}\frac{{{y}{\left.{d}{x}\right.}}}{{{3}{x}}}$

Then, divide both sides by y:

$\displaystyle\frac{{{\left.{d}{y}\right.}}}{{y}}={2}\frac{{{y}{\left.{d}{x}\right.}}}{{{3}{x}{y}}}$

$\displaystyle\frac{{{\left.{d}{y}\right.}}}{{y}}={2}\frac{{{\left.{d}{x}\right.}}}{{{3}{x}}}$

To be able to solve for the equation of the graph, we solve for the indefinite integral on both sides.

The problem becomes: $\displaystyle\int\frac{{{\left.{d}{y}\right.}}}{{y}}=\int{2}\frac{{{\left.{d}{x}\right.}}}{{{3}{x}}}$

For the left side,we integrate $\displaystyle\int\frac{{{\left.{d}{y}\right.}}}{{y}}$ using basic integration formula for logarithm: $\displaystyle\int\frac{{{d}{u}}}{{u}}={\ln}{\left|{u}\right|}+{C}$

$\displaystyle\int\frac{{{\left.{d}{y}\right.}}}{{y}}={\ln}{\left|{y}\right|}$

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

$\displaystyle\int{2}\frac{{{\left.{d}{x}\right.}}}{{{3}{x}}}={\left(\frac{{2}}{{3}}\right)}\int\frac{{{\left.{d}{x}\right.}}}{{{x}}}$

The integral part resembles the basic integration formula for logarithm: $\displaystyle\int\frac{{{d}{u}}}{{u}}={\ln}{\left|{u}\right|}+{C}$

$\displaystyle{\left(\frac{{2}}{{3}}\right)}\int\frac{{{\left.{d}{x}\right.}}}{{{x}}}={\left(\frac{{2}}{{3}}\right)}{\ln}{\left|{x}\right|}+{C}.$

Note: Just include the constant of integration "C" on one side as the arbitrary constant of a differential equation.

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

$\displaystyle{\ln}{\left|{y}\right|}={\left(\frac{{2}}{{3}}\right)}{\ln}{\left|{x}\right|}+{C}$

or $\displaystyle{y}={{e}}^{{{\left(\frac{{2}}{{3}}{\ln}{\left|{x}\right|}+{C}\right)}}}$

To solve for the equation of the graph that passes to a particular point $\displaystyle{\left({8},{2}\right)}$ , we plug-in $\displaystyle{x}={8}$ and $\displaystyle{y}={2}$ on the general solution: $\displaystyle{\ln}{\left|{y}\right|}={\left(\frac{{2}}{{3}}\right)}{\ln}{\left|{x}\right|}+{C}$ .

$\displaystyle{\ln}{\left|{2}\right|}={\left(\frac{{2}}{{3}}\right)}{\ln}{\left|{3}\right|}+{C}$

Isolate C:

$\displaystyle{C}={\ln}{\left|{2}\right|}-{\left(\frac{{2}}{{3}}\right)}{\ln}{\left|{3}\right|}$

Apply natural logarithm property: $\displaystyle{n}\cdot{\ln}{\left|{x}\right|}={\ln}{\left|{{x}}^{{n}}\right|}$ and $\displaystyle{\ln}{\left|{x}\right|}-{\ln}{\left|{y}\right|}={\ln}{\left|\frac{{x}}{{y}}\right|}$

$\displaystyle{C}={\ln}{\left|{2}\right|}-{\ln}{\left|{{3}}^{{\frac{{2}}{{3}}}}\right|}$

$\displaystyle{C}={\ln}{\left|\frac{{2}}{{{3}}^{{\frac{{2}}{{3}}}}}\right|}{\quad\text{or}\quad}{\ln}{\left|\frac{{2}}{{\sqrt[{{3}}]{{{9}}}}}\right|}$

Plug-in $\displaystyle{C}={\ln}{\left|\frac{{2}}{{\sqrt[{{3}}]{{{9}}}}}\right|}$ on the general solution: $\displaystyle{y}={{e}}^{{{\left(\frac{{2}}{{3}}{\ln}{\left|{x}\right|}+{C}\right)}}}$ , we get the equation of the graph that passes through (8,2) as:

$\displaystyle{y}={{e}}^{{{\left(\frac{{2}}{{3}}\right)}{\ln}{\left|{x}\right|}+{\ln}{\left|\frac{{2}}{{\sqrt[{{3}}]{{{9}}}}}\right|}}}$

Which simplifies to,

$\displaystyle{y}={{e}}^{{{\left(\frac{{2}}{{3}}\right)}{\ln{{\left({x}\right)}}}}}\cdot{{e}}^{{{\ln{{\left(\frac{{2}}{{\sqrt[{{3}}]{{{9}}}}}\right)}}}}}$

$\displaystyle{y}=\frac{{2}}{{\sqrt[{{3}}]{{{9}}}}}{{x}}^{{\frac{{2}}{{3}}}}$ as the final answer

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Friday, March 5, 2010

$\displaystyle{\left({8},{2}\right)},{y}'={2}\frac{{y}}{{{3}{x}}}$ Find an equation of the graph that passes through the point and has the given slope

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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?

Friday, March 5, 2010

Find an equation of the graph that passes through the point and has the given slope

No comments:

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?

$\displaystyle{\left({8},{2}\right)},{y}'={2}\frac{{y}}{{{3}{x}}}$ Find an equation of the graph that passes through the point and has the given slope

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