# Separable Variables

• Jan 31st 2009, 06:46 PM
element
Separable Variables
I'm currently having trouble with these two problems:
1. Solve the given differential equation by separation of variables
(e^x + e^-x) dy/dx = y^2

2. Find implicit and an explicit solution of the given IVP
sqr root of (1-y^2) dx - sqr root of (1-x^2) dy = 0 ; y(0) = sqr root 3 / 2
• Jan 31st 2009, 07:15 PM
Jhevon
Quote:

Originally Posted by element
I'm currently having trouble with these two problems:
1. Solve the given differential equation by separation of variables
(e^x + e^-x) dy/dx = y^2

note that we can write this as

$\displaystyle \frac 1{y^2}~dy = \frac 1{e^x + e^{-x}}~dx$

now integrate both sides to continue

Quote:

2. Find implicit and an explicit solution of the given IVP
sqr root of (1-y^2) dx - sqr root of (1-x^2) dy = 0 ; y(0) = sqr root 3 / 2
note that we can write the equation as

$\displaystyle \frac 1{\sqrt{1 - y^2}}~dy = \frac 1{\sqrt{1 - x^2}}~dx$

now integrate both sides to continue
• Jan 31st 2009, 07:21 PM
element
Quote:

Originally Posted by Jhevon
note that we can write this as

$\displaystyle \frac 1{y^2}~dy = \frac 1{e^x + e^{-x}}~dx$

now integrate both sides to continue

note that we can write the equation as

$\displaystyle \frac 1{\sqrt{1 - y^2}}~dy = \frac 1{\sqrt{1 - x^2}}~dx$

now integrate both sides to continue

I got that far for both 1 and 2. For 1, I'm stuck on integrating the right side. For 2, I got an answer but its not matching the books answer. Can you show me how you would do both? Thank you very much.
• Jan 31st 2009, 07:31 PM
Jhevon
Quote:

Originally Posted by element
I got that far for both 1 and 2. For 1, I'm stuck on integrating the right side.

note that $\displaystyle \int \frac 1{e^x + e^{-x}}~dx = \int \frac {e^x}{e^{2x} + 1}~dx$ .......(by multiplying by $\displaystyle \frac {e^x}{e^x}$)

now do a substitution of $\displaystyle u = e^x$

Quote:

For 2, I got an answer but its not matching the books answer. Can you show me how you would do both? Thank you very much.
integrating both sides we get

$\displaystyle \arcsin y = \arcsin x + C$

$\displaystyle \Rightarrow y = \sin (\arcsin x + C)$

now use your initial data to find the solution
• Jan 31st 2009, 08:11 PM
element
Thank you very much. Just one quick general question, when finding the implicit and explicit solution, are we to always have "y" by itself?
• Jan 31st 2009, 08:23 PM
Jhevon
Quote:

Originally Posted by element
Thank you very much. Just one quick general question, when finding the implicit and explicit solution, are we to always have "y" by itself?

"explicit" means y by itself, "implicit" means x's and y's mixed, or simply, not y by itself.