1. ## Differential Equation help...

Hello, I'm a newbie when it comes to differential equations and my professor assigned some problems that we never went over in class. Hopefully someone can help...

y' = 2y; y(x) = Ce^(2x), y(0) = 3.

Here are the directions: First verify that y(x) satisfies the given differential equation. Then determine a value of the constant C so that y(x) satisfies the given initial condition.

I have to do about 10 of these, so if anyone could go through the steps on how to solve this, I will greatly appreciate it. Thanks for your time...

2. $\displaystyle \frac{dy}{dx} = 2y$

$\displaystyle \frac{1}{y}\,\frac{dy}{dx} = 2$

$\displaystyle \int{\frac{1}{y}\,\frac{dy}{dx}\,dx} = \int{2\,dx}$

$\displaystyle \int{\frac{1}{y}\,dy} = 2x + C_1$

$\displaystyle \ln{|y|} + C_2 = 2x + C_1$

$\displaystyle \ln{|y|} = 2x + C_1 - C_2$

$\displaystyle |y| = e^{2x + C_1 - C_2}$

$\displaystyle |y| = e^{C_1 - C_2}e^{2x}$

$\displaystyle y = \pm e^{C_1 - C_2}e^{2x}$

$\displaystyle y = Ce^{2x}$ where $\displaystyle C = \pm e^{C_1 - C_2}$.

Now use the inital condition $\displaystyle y(0) = 3$

$\displaystyle 3 = Ce^{2\cdot 0}$

$\displaystyle 3 = C$.

Therefore $\displaystyle y = 3e^{2x}$.