# Separable Equation ..

• Feb 19th 2010, 11:33 AM
Ted
Separable Equation ..
Hello.

Problem: Find the solution of the differenial equation that satisfies the given intial condition.
$\displaystyle \left( 2y + e^{3y} \right) \frac{dy}{dx} = x cos(x)$
The intial condition: $\displaystyle y(0)=0$.

My Solution:

I rewrite it as:

$\displaystyle \left( 2y + e^{3y} \right)dy=xcos(x)dx$

By integrating both sides and using integration by parts for the right side, I obtained:

$\displaystyle y^2+\frac{1}{3}e^{3y}=xsin(x)+cos(x)+C$.

The problem here is that I can not solve the last equation for $\displaystyle y$ so that I can use the intial condition to find the particular solution.
Any help?
• Feb 19th 2010, 11:42 AM
pickslides
Quote:

Originally Posted by Ted
The problem here is that I can not solve the last equation for $\displaystyle y$ so that I can use the intial condition to find the particular solution.
Any help?

You can use the initial condition at this point to find $\displaystyle c$. You just won't be able to put it into the form $\displaystyle y = f(x)$
• Feb 19th 2010, 11:45 AM
Ted
Quote:

Originally Posted by pickslides
You can use the initial condition at this point to find $\displaystyle c$. You just won't be able to put it into the form $\displaystyle y = f(x)$

Hmmm.
OK I will put $\displaystyle x=0$.
• Feb 19th 2010, 11:47 AM
pickslides
Quote:

Originally Posted by Ted
The intial condition: $\displaystyle y(0)=0$.

This means when $\displaystyle x=0$ then $\displaystyle y = 0$
• Feb 19th 2010, 01:23 PM
Ted
Quote:

Originally Posted by pickslides
This means when $\displaystyle x=0$ then $\displaystyle y = 0$

Sorry I did not get this one.
What I know is: $\displaystyle y(0)=0$ solve the equation for $\displaystyle y$, then in the resulting function put $\displaystyle x=0$ and equate it to zero, and solve the resulting equation for the constant and substitute the value of the constant in the general solution to get the particular solution.
I did not understand why this means I should put $\displaystyle y=0$.

$\displaystyle y$ is assumed to be a function of $\displaystyle x$, i.e. $\displaystyle {\color{red}y} = y({\color{blue}x})$.
The initial condition says that $\displaystyle y({\color{blue}0}) = {\color{red}0}$. This means that your particular solution has to pass through the point $\displaystyle ({\color{blue}x},{\color{red}y}) = ({\color{blue}0},{\color{red}0})$ .
You don't have to solve for $\displaystyle y$ explicitly. All you have to do is plug in $\displaystyle x=0$ and $\displaystyle y=0$ and solve for $\displaystyle C$.
So: $\displaystyle ({\color{red}0})^2+\frac{1}{3}e^{3({\color{red}0}) } = {\color{blue}0}\sin ({\color{blue}0}) + \cdots$ etc.