Need Help with Solving an ODE in Calculus 1?

[schapman22]

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

the answer is supposed to be X(t). Thank you in advance.

 

[TylerH]

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.

 

[schapman22]

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?

 

[schapman22]

*x = 2.828t

 

[TylerH]

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...

 

[schapman22]

how would i do the integral of the left side?

 

[lurflurf]

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

 

[schapman22]

actually no

 

[omkar13]

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)

 

[TylerH]

\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]

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.

 

[schapman22]

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