Separable Equation ..

**Ted** · February 19th 2010, 11:33 AM

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?

**pickslides** · February 19th 2010, 11:42 AM

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 $y = f(x)$

**Ted** · February 19th 2010, 11:45 AM

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 $x=0$ .
What about the y's ?

**pickslides** · February 19th 2010, 11:47 AM

Originally Posted by Ted

The intial condition: $y(0)=0$ .

This means when $x=0$ then $y = 0$

**Ted** · February 19th 2010, 01:23 PM

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 $y=0$ .

please, anyone can explain it?

**o_O** · February 19th 2010, 01:30 PM

$y$ 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.

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