# Separable Equation With Intial Condition.

• Feb 19th 2010, 09:26 AM
Ted
Separable Equation With Intial Condition.
Hello.

Problem: Find the solution of the differential equation that satisfies the given intial condition.
$\frac{dL}{dt}=kL^2ln(t)$
The intial condition: $L(1)=-1$.

My solution:

Well, I noticed its a separable equation,
The problem here, is the question did not identify the K!
so I assumed its a constant.
then I rewrite the D.E. as:

$\frac{dL}{L^2}=k ln(t) dt$

By integrating both sides and using integration by parts, I obtained:

$\frac{-1}{L}=k ( tln(t)-t+C)$

$\frac{-1}{L}=k ( tln(t)-t)+A$ where $A=Ck$.

Solving for L,I got:

$L=\frac{-1}{k(tln(t)-t)+A}$.

since $L(1)=-1$:

$L(1)=\frac{-1}{-k+A}=-1$.

so:

$A-k=1$, i.e. $A=1+k$.

Then ????
I stopped here!
• Feb 19th 2010, 10:36 AM
e^(i*pi)
From what I can gather you've gone as far as you can without knowing the value of k.

If I were doing this question I'd leave the equation for L in terms of t and k
• Feb 19th 2010, 10:51 AM
Ted
So the particular solution is:

$L=\frac{-1}{k(tln(t)-t)+k+1}$

Right?
• Feb 19th 2010, 10:58 AM
e^(i*pi)
Quote:

Originally Posted by Ted
So the particular solution is:

$L=\frac{-1}{k(tln(t)-t)+k+1}$

Right?

That's what I'd put, it's possible you might be given k but since A is pretty much the constant of integration it'd be better to have the equation in terms of k