1. ## seperating variables

If dy/dx = -10y and if y = 50 when x = 0, then y = (a) 50cos10x (b) 50e^(10x) (c) 50e^(-10x) (d) 50 - 10x (e) 50 - 5x^2

2. Originally Posted by LeoBloom.
If dy/dx = -10y and if y = 50 when x = 0, then y = (a) 50cos10x (b) 50e^(10x) (c) 50e^(-10x) (d) 50 - 10x (e) 50 - 5x^2

Notice this is a seperable equation

$\frac{dy}{dx}=-10y$

Getting all of the y's on one side gives

$\frac{1}{y}dy=-10dx$

Now you integrate both sides

$\ln|y|=-10x+C$

Now solving for y

$y=e^{-10x+C}=e^{-10x}e^{C}=ke^{-10x}$ where $k=\pm e^{C}$

Now using the intial condition you can solve for k.

$50=ke^{-10(0)}$

$50=k$

So the solution to the differential equation is

$y=50e^{-10x}$

3. Have you come across the technique of separating variables yet?

$
\frac{dy}{dx}=-10y$

divide through by y and integrate both sides with respect to x
$
\int \frac{1}{y} \frac{dy}{dx} dx=\int -10 dx$

giving us
$
\int \frac{1}{y}dy=\int -10 dx$

do the integration leaving us with
$ln |y| = -10x+c$

I'll leave you to finish it off and plug in the boundary condition

4. Right, thanks alot guys, yes, we have done this before in class but for some reason it flew out of my head.