- #1
erok81
- 464
- 0
Homework Statement
dx/dt=3x(x-5); x(0)=2
Homework Equations
dx/dt=F(x)G(t)
dx/dt=x'
The Attempt at a Solution
First I separated the variables using F=3x(x-5) and G=3
x'/(3x(x-5))=3
Applying method of quadrature:
[tex]\int\frac{x'}{(3x(x-5)}dt=\int3 dt[/tex]
On the right hand side I end up with 3t. For the left hand side I get an integral I need apply partial fractions.
After solving the partial fraction I have:
[tex]\int\frac{1}{5x-25}-\frac{1}{5x}[/tex]
Integrating that leaves me with:
1/5ln|5x-25|-1/5ln|5x|=3t+c
Multiplying both sides by 5:
ln|5x-25|-ln|5x|=15t+c
Applying log property where ln|a|-ln|b|=ln|a/b|
[tex]ln\frac{5x-25}{5x}[/tex]
Reducing that by 5:
[tex]ln\frac{x-5}{x}[/tex]
So finally I have...after taking e of both sides:
[tex]\frac{x-5}{x}=e^{15t}+c[/tex]
I am kind of stuck here solving for c...assuming I've done the above correctly.
The initial condition shows x(0)=2.
I've tried sub'ing in x=2 and t=0 to no avail. I've either done something wrong at the top or messed up this last part.
Any help will be appreciated.