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

The given slope equation: `y' =2y/(3x)` is in form of first order ordinary differential equation. In order to evaluate this, we let `y'`  as `(dy)/(dx)` .

`(dy)/(dx)=2y/(3x)`

Then, express as a variable separable differential equation: `N(y) dy= M(x) dx` .

To accomplish this, we cross-multiply `dx` to the...

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The given slope equation: `y' =2y/(3x)` is in form of first order ordinary differential equation. In order to evaluate this, we let `y'`  as `(dy)/(dx)` .

`(dy)/(dx)=2y/(3x)`

Then, express as a variable separable differential equation: `N(y) dy= M(x) dx` .

To accomplish this, we cross-multiply `dx` to the other side.

`dy=2(ydx)/(3x)`

Then, divide both sides by y:

`(dy)/y=2(ydx)/(3xy)`

`(dy)/y=2(dx)/(3x)`

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

The problem becomes: `int(dy)/y= int 2(dx)/(3x)`

For the left side,we integrate `int(dy)/y` using basic integration formula for logarithm: `int (du)/u = ln|u|+C`

`int (dy)/y = ln|y|`

For the right side, we may apply basic integration property: `int c*f(x)dx =c intf(x)dx` .

`int 2(dx)/(3x)=(2/3)int (dx)/(x)`

The integral part resembles the basic integration formula for logarithm:` int (du)/u = ln|u|+C`

`(2/3)int (dx)/(x)=(2/3)ln|x|+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:

`ln|y|=(2/3)ln|x|+C`

or `y = e^((2/3ln|x|+C))`

To solve for the equation of the graph that passes to a particular point `(8,2)` , we plug-in `x=8` and` y =2` on the general solution: `ln|y|=(2/3)ln|x|+C` .

`ln|2|=(2/3)ln|3|+C`

Isolate C:

`C =ln|2|-(2/3)ln|3|`

Apply natural logarithm property: `n*ln|x|= ln|x^n|` and `ln|x|-ln|y| = ln|x/y|`

`C =ln|2|-ln|3^(2/3)|`

`C=ln|2/3^(2/3)| orln|2/root(3)(9)|`

Plug-in `C=ln|2/root(3)(9)|` on the general solution: `y = e^((2/3ln|x|+C))` , we get the equation of the graph that passes through (8,2) as:

`y = e^((2/3)ln|x|+ln|2/root(3)(9)|)`

Which simplifies to,

`y = e^((2/3)ln(x))*e^(ln(2/root(3)(9)))`

`y = 2/root(3)(9)x^(2/3)` as the final answer

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