Question

Solve x(t) for 5x’(t)+3x(t)=6, x(0)=1, where x’(t) is the derivative of x(t) with respect to t.

Answer 1

put it in standard form and use integrating factor

Answer 2

or wolframalpha

Answer 3

im confuse

Answer 4

Try these http://www.khanacademy.org/#differential-equations

Answer 5

x'(t) + (3/5)x(t)= 6/5
x'(t) + (3/5)x(t) - 6/5= 0
(1/2)x(t) + (3/5)x(t) - 6/5= 0
(1/2)x(t) + (3/5)x(t) - 6/5= 0 ??

Answer 6

plpeter673 that's not correct,which grade you are in high school?

Answer 7

y

Answer 8

who's asking

Answer 9

I'll help you according to your grade , can be solved in several ways

Answer 10

yes, hs

Answer 11

okay, Do you know what's integrating factor?

Answer 12

no

Answer 13

we have
\[5x'(t)+3x(t)=6\]Let's write this as
\[ 5\frac{dx}{dt}+3 x=6\]
where x= x(t)
\[ 5 \frac{dx}{dt}= 6-3x\]
This can be written as
\[ \frac{dx}{6-3x}= \frac{dt}{5}\]
now integrate both sides
We know integration of \(\frac{1}{a-x}\)= \(-\ln(a-x)\) here ln= log to the base e
We get
\[ -\frac{1}{3} \times \ln(6-3x) = \frac{t}{5}+C\]
C= constant of integration
We get
\[ \ln {6-3x} = -\frac{3t}{5} +C\]
Take antilog both sides
\[ 6-3x= e^{-\frac{3t}{5}}\times C\]
we get x(t)
\[x(t)= \frac{6-C\times e^{-\frac{3t}{5}}}{3} \]
we are given at t=0 x(t)=1
so
\[1=\frac{6-C\times1}{3} \]
\[3=6-C\]
we get
\[C= 3\]
so
\[x(t)= \frac{6-3\times e^{-\frac{3t}{5}}}{3} \]

Answer 14

x(t) can also be written as
\[x(t)=2-e^{-\frac{3t}{5}}\]

Answer 15

cool. thk

Answer 16

welcome :) Did you get this?

Answer 17

yes