# Find the unique solution to y'+y/t*lnt=9t^2, t>0, y(e)=2e^3.

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### 1 Answer

You should use integrating factor method such that:

`P(t) = 1/(t*lnt), hence the integarting factor isI(t) = e^int_e^t (ds)/(s*lns) = e^(ln (ln t) - ln(ln e)) = e^(ln (ln t) - ln(1)) = e^(ln (ln t)) = ln t`

`` Multiplying the equation by integrating factor I(t) yields:

`y'*ln t + (y*ln t)/t*lnt=9t^2*ln t =gt y'*ln t + y/t=9t^2*ln t`

Notice that `y'*ln t + y/t = y'*ln t + y*(ln t)' = (I(t)*y)'`

Hence,`(I(t)*y)' = 9t^2*I(t)`

Integrating both sides yields:

`int_e^t (I(t)*y)' = int_e^t 9t^2*I(t) dt`

`` `I(t)*y(t) - I(e)*y(e) = int_e^t 9t^2*I(t) dt`

You need to use integration by parts to solve`int_e^t 9t^2*I(t) dt` .

You should come up with the notation:

`u = ln t =gt du = (dt)/t`

`dv = 9t^2 =gt v = 3t^3`

`int udv = uv - int vdu=gt int_e^t 9t^2 ln t dt = 3t^3*ln t|_e^t - int_e^t (3t^3dt)/t`

`int_e^t 9t^2 ln t dt = 3t^3*ln t - 3e^3 - t^3 + e^3`

`int_e^t 9t^2 ln t dt = 3t^3*ln t - 2e^3 - t^3`

You need to evaluate `I(e)*y(e) = ln e*2e^3 = 2e^3`

Hence, `I(t)*y(t) - I(e)*y(e) = y(t)*ln t - 2e^3`

`y(t)*ln t - 2e^3 = 3t^3*ln t - 2e^3 - t^3`

Reducing by - `2e^3` both sides yields:

`y(t)*ln t = 3t^3*ln t - t^3`

`` `y(t) = 3t^3 - t^3/ln t`

**Hence, evaluating the unique solution to differential equation yields `y(t) = 3t^3 - t^3/ln t` .**