Find the solution of the differential equation that satisfies the given initial condition. `(dL)/(dt) = kL^2 ln(t)` L(1) = -1

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Chapter 9, 9.3 - Problem 18 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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ishpiro | College Teacher | (Level 1) Educator

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This differential equation can be solved by the method of separating variables. 

Multiply both sides by dt and divide both sides of the equation by `L^2`  . This will bring all terms depending on t to the right side and all terms depending on L to the left side:

`(dL)/L^2 = kln(t)dt`

Now both sides can be integrated. On the left side, we have an integral of a power function:

`int (dL)/L^2 = int L^(-2) dL = -1L^(-1) = -1/L`

(The constant of integration can be omitted for now.)

On the right side, the integral of ln(t) equals

`int ln(t)dt = tln(t) - t`

(This can be verified by taking the derivative using the product rule and the fact that (ln(t))' = 1/t : (tln(t) - t)' = ln(t) + t/t - 1 = ln(t))

So the after the integration the equation becomes

`-1/L = k(tln(t) - t) + C`

The constant of integration C is added to the right side here.

Since the initial condition is given, L(1) = -1, C can be found now:

for t = 1, ln(t) = ln(1) = 0, and

`-1/(-1) = k(1*0 - 1) + C = -k + C`

1 = -k + C, so from here C = 1 + k

Putting C back into the equation, we get

`-1/L = k(tln(t)-t) + 1 + k`

Multiplying both sides by -1 and opening parenthesis on the right side results in

`1/L = -ktln(t) + kt - 1 - k`

From here,

`L(t) = 1/(kt - ktln(t) - 1 - k)`

So, the solution of the given equation satisfying initial condition L(1) = -1 is

`L(t) = 1/(kt-kln(t)-1-k)`

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