1. ## Differential Equation (ii)

I'm stuck at this question.

Express G in terms of t and k.

$\displaystyle \frac{dG}{dt}=\frac{25k-G}{25}$

ANSWER: $\displaystyle G=25k(1-e^{\frac{-t}{25}})$

2. We can use an integrating factor.

$\displaystyle \frac{dG}{dt}=k-\frac{G}{25}$

$\displaystyle \frac{dG}{dt}+\frac{G}{25}=k$

The integrating factor is

$\displaystyle e^{\int\frac{1}{25}dt}=e^{\frac{t}{25}}$

$\displaystyle \frac{d}{dt}[e^{\frac{t}{25}}G]=ke^{\frac{t}{25}}$

Integrate:

$\displaystyle Ge^{\frac{t}{25}}=25ke^{\frac{t}{25}}+C$

$\displaystyle \boxed{G=25k+C_{1}e^{\frac{-t}{25}}}$

3. Originally Posted by r_maths
I'm stuck at this question.

Express G in terms of t and k.

$\displaystyle \frac{dG}{dt}=\frac{25k-G}{25}$

ANSWER: $\displaystyle G=25k(1-e^{\frac{-t}{25}})$
Separate the variables:

$\displaystyle \frac{dG}{25k-G}=\frac{dt}{25}$

Integrate:

$\displaystyle -ln{\left|25k-G\right|}=\frac{t}{25}+C$.

Solve for G:

$\displaystyle ln{\left|25k-G\right|}=-\frac{t}{25}+C$
$\displaystyle {\left|25k-G\right|}=e^{-\frac{t}{25}+C}$
$\displaystyle {\left|25k-G\right|}=e^{C}e^{-\frac{t}{25}}$
$\displaystyle 25k-G=\pm e^{C}e^{-\frac{t}{25}}$
$\displaystyle G=25k-Ce^{-\frac{t}{25}}$

Was there an initial condition? $\displaystyle G(0)=0$?

$\displaystyle 0=25k-Ce^{-\frac{0}{25}}\rightarrow C=25k$

$\displaystyle \therefore G=25k-25ke^{-\frac{t}{25}}$
$\displaystyle \color{red}\boxed{G=25k{\left(1-e^{-\frac{t}{25}}\right)}}$

Hope that helped you out!