# Simple DE....but lack of practice.

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• Aug 28th 2010, 05:43 AM
JavaJunkie
Simple DE....but lack of practice.
Hey,
I haven't looked at differential equations for some years... and I'm having trouble with two really simple ones....(Speechless)

$\displaystyle \frac{dx}{dt} = (x + 2), x(0) = 1$ ........ODE (1)

and $\displaystyle \frac{dy}{dt} = 2y, y(0) = 1$............ ODE (2)

The text I'm looking at says the answers for ODE (1) and ODE (2) should be $\displaystyle x + 2 = 3e^t$ and $\displaystyle y = e^{2t}$ respectively.

I thought the solutions (not imposing the initial conditions are) for each of the conditions are

$\displaystyle xt + 2t$ and $\displaystyle 2yt$ respectively.

I must be doing this incorrectly. Do I use separation of variables?

$\displaystyle \int \frac{dx}{x+2} = \int \ dt$

and

$\displaystyle \int \frac{dy}{y} = \int 2 dt$ ?

If so, how do I integrate $\displaystyle \int \frac{dx}{x+2}$ and $\displaystyle \int \frac{dy}{y}$ ?

Thanks
• Aug 28th 2010, 06:33 AM
Sudharaka
Quote:

Originally Posted by JavaJunkie
Hey,
I haven't looked at differential equations for some years... and I'm having trouble with two really simple ones....(Speechless)

$\displaystyle \frac{dx}{dt} = (x + 2), x(0) = 1$ ........ODE (1)

and $\displaystyle \frac{dy}{dt} = 2y, y(0) = 1$............ ODE (2)

The text I'm looking at says the answers for ODE (1) and ODE (2) should be $\displaystyle x + 2 = 3e^t$ and $\displaystyle y = e^{2t}$ respectively.

I thought the solutions (not imposing the initial conditions are) for each of the conditions are

$\displaystyle xt + 2t$ and $\displaystyle 2yt$ respectively.

I must be doing this incorrectly. Do I use separation of variables?

$\displaystyle \int \frac{dx}{x+2} = \int \ dt$

and

$\displaystyle \int \frac{dy}{y} = \int 2 dt$ ?

If so, how do I integrate $\displaystyle \int \frac{dx}{x+2}$ and $\displaystyle \int \frac{dy}{y}$ ?

Thanks

Dear JavaJunkie,

Yes you should use seperation of variables. And remember that, $\displaystyle \int{\frac{1}{x+c}}dx=\ln{\mid{x+c}\mid}~;~where~C ~is~an~arbitary~constant.$ Hope you would be able to continue.