# Trouble solving this ode

I am having trouble solving this ode, I wasn't sure if this should go under calculus section or differential equations section, but I figured since were given this in calculus 1 it belongs here. We just recently started learning integral calculus and I don't know a whole lot about differential equations. How would you solve this?

dx/dt = 2000-500x/100

Get all the x's to the dx/dt side, then integrate both sides. When I say integrate, I mean integrate the side with the x's in terms of x and integrate the other in terms of t. That will leave with something that you can use simple algebraic manipulation to get it into a explicit form.

When i tried that i got
dx/dt - 5x = 20
dx - 5x = 20 dt
∫dx - 5x = ∫20 dt
(5x^2)/2 = 20t
5x^2 = 40t
x^2 = 8
x = 2.828

am I on the right track here?

*x = 2.828t

Oh, I forgot to mention that you have to get some function of x to be multiplied by dx. In other words, dx/x is okay, but dx+x isn't. More generally, f(x)dx is okay, f(x)+dx isn't.

$$\frac{dx}{dt}=2000-5x=5(400-x)$$
$$\frac{\frac{dx}{dt}}{400-x}=5$$
$$\int \frac{dx}{400-x}=\int 5 dt$$
and so on...

how would i do the integral of the left side?

lurflurf
Homework Helper
Have you heard of a function called log(x)?

actually no

dx/dt = 2000-500x/100
=>dx/dt=20-5x
=>dx/20-5x=dt=>(dx/4-x)=5dt
Integrating it,-ln(4-x)=5t+k(integrating constant)
=>x=4-a*exp(-5t)
(ln(x) is one of the basic functions in maths)

$$\int \frac{dx}{400-x}$$ is a specific case of $$\int \frac{dx}{x}$$. Given $$\int \frac{dx}{x} = ln(x)$$, you can solve it using substitution.

BTW, omkar13 is close but wrong. The correct answer can be found on Wolfram Alpha, if you want proof: http://www.wolframalpha.com/input/?i=dx/dt+=+2000-500x/100.

HallsofIvy
Homework Helper
When i tried that i got
dx/dt - 5x = 20
dx - 5x = 20 dt
This is just bad algebra. If you start with dx/dt- 5x= 20 and multiply both sides by dt, you get dx- 5xdt= 20dt. However, the best thing to do is to go back to dx/dt= 5x+ 20, multiply both sides by dt and divide both sides by 5x+ 20: dx/(5x+20)= dt. Now let u= 5x+ 20 so that du= 5dx or dx= (1/5)du. The equation becomes (1/5) du/u= dt. Integrat both sides, then go back to x.

∫dx - 5x = ∫20 dt
(5x^2)/2 = 20t
No, "dx- 5x" (which was wrong anyway) is NOT -5xdx

5x^2 = 40t
x^2 = 8
x = 2.828

am I on the right track here?
Pretty much every thing you did, both algebra and Calculus, is wrong.

dude we just started on the integral in class you don't have to be douchebag I was just asking for help