# Differential Equation question?

dL/dt = kL^(2)*ln(t) , L(1)=-9

PLEASE HELP!

### 2 Answers

- rotchmLv 73 months ago
Hint: You Are not really trying, arent you!? Since most of this is just highschool algebra, something you learned many years ago.

Again, using highschool algebra, rewrite the DE in a way that all the variables L are on the left and the t's on the right. This is also called a separable DE...We have shown you this in your previous post. What do you get when you do that?

Answer that and then we will proceed...

- 3 months ago
dL / L^2 = k * ln(t) * dt

Integrate

-1/L = k * (t * ln(t) - t) + C

L = -9 when t = 1

-1/(-9) = k * (1 * ln(1) - 1) + C

1/9 = -k + C

k + 1/9 = C

(9k + 1) / 9 = C

-1/L = k * (t * ln(t) - t) + (1/9) * (9k + 1)

-1/L = (9k * (t * ln(t) - t) + 9k + 1) / 9

-L = 9 / (9k * t * (ln(t) - 1) + 9k + 1)

L = 9 / (9k * t * (1 - ln(t)) - 9k - 1)

L = 9 / (9k * (t - t * ln(t) - 1) - 1)

L = 9 / (9k * (t * (1 - ln(t)) - 1) - 1)