Question

help plsssss... medals and fan
solve dx/dt = 0.05 (20-x)

Answer 1

@sammixboo help pls ?

Answer 2

@iGreen

Answer 3

@IrishBoy123

Answer 4

@dan815

Answer 5

@yajna

how did you try it? useful to know how experienced you are at this.

Answer 6

by integration, differential

Answer 7

@jim_thompson5910

Answer 8

\[\Large \frac{dx}{dt} = 0.5(20-x)\]

\[\Large dx = 0.5(20-x)*dt\]

\[\Large \frac{dx}{20-x} = 0.5dt\]

\[\Large \int\frac{dx}{20-x} = \int 0.5dt\]

I'll let you finish up

Answer 9

yeah i ve reached so far,, i am having issues

Answer 10

hint: let u = 20 - x, so du = -dx which means dx = -du

Answer 11

\[\Large \int\frac{dx}{20-x} = \int 0.5dt\]

\[\Large \int\frac{1}{20-x}dx = \int 0.5dt\]

\[\Large \int\frac{1}{u}(-du) = \int 0.5dt\]

\[\Large -\int\frac{1}{u}du = \int 0.5dt\]

Answer 12

so far i have done it right

Answer 13

ok what's next?

Answer 14

@Michele_Laino

Answer 15

what part you need help on?

Answer 16

and up to what do you have?

Answer 17

I HAVE TO INTEGRATE IT

Answer 18

yep
can you do that?

Answer 19

these integrals you have are pretty easy to evaluate (no sub needed )
\[\int\limits_{}^{}\frac{1}{x} dx=\ln|x|+C \\ \int\limits_{}^{}c dx=cx+K\]

Answer 20

WHAT about 1/(20-x) ?

Answer 21

@jim_thompson5910 took the liberty of doing the sub for you

Answer 22

you do know that if u=20-x
then du/dx=-1
or du=- dx
right?

Answer 23

that is how he got the integral on the left hand side