Differential Equation (ii)

**r_maths** · May 10th 2008, 09:56 AM

I'm stuck at this question.

Express G in terms of t and k.

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

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

**galactus** · May 10th 2008, 10:05 AM

We can use an integrating factor.

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

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

The integrating factor is

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

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

Integrate:

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

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

**Chris L T521** · May 10th 2008, 10:22 AM

Originally Posted by r_maths

I'm stuck at this question.

Express G in terms of t and k.

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

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

Separate the variables:

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

Integrate:

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

Solve for G:

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

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

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

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

Hope that helped you out!

Thread: Differential Equation (ii)

LinkBack

Thread Tools

Display

Differential Equation (ii)

Similar Math Help Forum Discussions

[SOLVED] Partial Differential Equation satisfy corresponding equation

Transforming a differential equation into another differential equation

problem on matrix ricaati equation for nonlinear singular differential equation

[SOLVED] Solve Differential equation for the original equation

Partial differential equation-wave equation - dimensional analysis

Search Tags