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# what is the general solve ?

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y'=(y(t+1)+(t+1)^2)/t^2

Jun 18, 2020
edited by Guest  Jun 18, 2020

#1
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Hint: Try to factoring out (t+1)

Jun 18, 2020
#2
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$$y' = \dfrac{y(t + 1) + (t + 1)^2}{t^2}\\ \dfrac{dy}{dt} = y\left(\dfrac1t + \dfrac1{t^2}\right) + \left(1 + \dfrac1t\right)^2\\ \dfrac{dy}{dt} - \left(\dfrac1t + \dfrac1{t^2}\right)y = \left(1 + \dfrac1t\right)^2\\$$

We then find the integrating factor:

$$\mathcal{I}(t) = e^{-\int \left(\frac1t + \frac1{t^2}\right)\,dt} = e^{-\ln t + \frac1t} = \dfrac1t e^{1/t}$$

This means

$$(y\mathcal I)' = \left(1 + \dfrac1t\right)^2 \mathcal I$$

$$\dfrac{y}t e^{1/t} = \displaystyle \int \dfrac1t \left(1 + \dfrac1t\right)^2 e^{1/t}\,dt$$

You will need some special functions to evaluate this integral, so good luck. Hope this helps.

Jun 18, 2020