seperating variables

**LeoBloom.** · May 20th 2008, 06:26 AM

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

**Jen** · May 20th 2008, 06:36 AM

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

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

**thelostchild** · May 20th 2008, 06:36 AM

Have you come across the technique of separating variables yet?

for your problem we have
$<br /> \frac{dy}{dx}=-10y$

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

giving us
$<br /> \int \frac{1}{y}dy=\int -10 dx$

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

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

**LeoBloom.** · May 20th 2008, 06:39 AM

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

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