1. ## Separable Equation ..

Hello.

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

My Solution:

I rewrite it as:

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

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

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

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

2. Originally Posted by Ted
The problem here is that I can not solve the last equation for $y$ 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 $c$. You just won't be able to put it into the form $y = f(x)$

3. Originally Posted by pickslides
You can use the initial condition at this point to find $c$. You just won't be able to put it into the form $y = f(x)$
Hmmm.
OK I will put $x=0$.

4. Originally Posted by Ted
The intial condition: $y(0)=0$.
This means when $x=0$ then $y = 0$

5. Originally Posted by pickslides
This means when $x=0$ then $y = 0$
Sorry I did not get this one.
What I know is: $y(0)=0$ solve the equation for $y$, then in the resulting function put $x=0$ 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 $y=0$.

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