# separation of variables

• Oct 9th 2008, 07:50 AM
2clients
separation of variables
Hello!

In this problem, find the equation of the graph which passes through the given point and has the indicated slope:

m = y' = dy/dx = (2y)/(3x)

(1,1)

So, after cross multiplying to separate, I have:

(1/2y) (dy) = (1/3x) (dx)

After integrating both sides:

Ln(y)/2 = Ln(x)/3

Simplifying:

Ln(y) = (2Ln(x))/3

Is this the right track? I can't remember how to get rid of the Ln on the left side to solve for y. How do I use the point (1,1) ?

Thank you so much!!
• Oct 9th 2008, 09:15 AM
Chris L T521
Quote:

Originally Posted by 2clients
Hello!

In this problem, find the equation of the graph which passes through the given point and has the indicated slope:

m = y' = dy/dx = (2y)/(3x)

(1,1)

So, after cross multiplying to separate, I have:

(1/2y) (dy) = (1/3x) (dx)

After integrating both sides:

Ln(y)/2 = Ln(x)/3

Simplifying:

Ln(y) = (2Ln(x))/3

Is this the right track? I can't remember how to get rid of the Ln on the left side to solve for y. How do I use the point (1,1) ?

Thank you so much!!

Yes, you're on the right track, but when you integrate both sides, don't forget to throw the constant $\displaystyle C$ on the side with the x's. So you should end up with

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

Now use the point as the initial condition: $\displaystyle (1,1)\implies y(1)=1$

Thus, $\displaystyle 1=C(1)^{\frac{2}{3}}\implies C=1$

So the final answer would be $\displaystyle \color{red}\boxed{y=x^{\frac{2}{3}}}$

Does this make sense?

--Chris
• Oct 9th 2008, 09:25 AM
2clients
Thank you, Chris.

It does make sense, mostly - I just can't reason why the C only goes on the one side?
• Oct 9th 2008, 09:29 AM
Chris L T521
Quote:

Originally Posted by 2clients
Thank you, Chris.

It does make sense, mostly - I just can't reason why the C only goes on the one side?

It really appears on both sides after integration:

$\displaystyle \tfrac{1}{2}\ln y+C_1=\tfrac{1}{3}\ln x +C_2$

Solving for y, you end up with $\displaystyle y=e^{\frac{2}{3}\ln x+{\color{blue}C_2-C_1}}$

But $\displaystyle C_2-C_1$ is just another constant; let's call it $\displaystyle C$

So, $\displaystyle y=e^{\frac{2}{3}\ln x+{\color{blue}C}}=\dots$

Does this clarify things?

--Chris
• Oct 9th 2008, 09:34 AM
2clients
yes perfectly, thanks again!