- #1

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dx/dt = 2000-500x/100

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

- Thread starter schapman22
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- #1

- 74

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dx/dt = 2000-500x/100

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

- #2

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- #3

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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?

- #4

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*x = 2.828t

- #5

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[tex]\frac{dx}{dt}=2000-5x=5(400-x)[/tex]

[tex]\frac{\frac{dx}{dt}}{400-x}=5[/tex]

[tex]\int \frac{dx}{400-x}=\int 5 dt[/tex]

and so on...

- #6

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how would i do the integral of the left side?

- #7

lurflurf

Homework Helper

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

- #8

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actually no

- #9

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=>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)

- #10

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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.

- #11

HallsofIvy

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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.When i tried that i got

dx/dt - 5x = 20

dx - 5x = 20 dt

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

(5x^2)/2 = 20t

Pretty much every thing you did, both algebra and Calculus, is wrong.5x^2 = 40t

x^2 = 8

x = 2.828

am I on the right track here?

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