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Show that there is a unique solution of the differential equation $tx'(t)=(2t^2+1)x(t)+t^2,t>0$, which has a finite limit as $t\rightarrow \infty$ and compute this limit.

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1 Answer 1

Rewrite $$ tx' - (2t^2 + 1)x = t^2 $$ Multiply through by integration factor, $\mu(t)$, and divide by $t$: $$ \mu x' - (2t + 1/t)\mu x = t\mu \;\;\;\;\;(1)$$

Define $\mu(t)$ such that: $-(2t+1/t)\mu = \mu'$ (2).

Plugging (2) into (1) we get: $$ \mu x' + \mu' x = t\mu$$ Reverse the product rule: $$ \frac{d}{dt} (\mu x) = t\mu$$

Thus, $\mu x = \int t\mu dt$ and finally: $$ x(t) = \frac{\int t\mu(t) dt}{\mu(t)} \;\;\;\;\;\;\;(3) $$

From (2) we see that $-(2t + 1/t) = \mu'/\mu = d/dt\ln(\mu)$ and thus: $$ \ln(\mu) = \int -(2t + 1/t) \;dt = -t^2 - \ln(t) + C $$ Solve for $\mu(t)$: $$ \mu(t) = e^{-t^2 - \ln(t) + C} = Ce^{-t^2}/t \;\;\;\;\;(4)$$

Plug (4) into (3): $$ x(t) = \frac{\int t Ce^{-t^2}/t \;dt}{Ce^{-t^2}/t} = \frac{\int e^{-t^2} \;dt}{e^{-t^2}/t}$$

The right hand integral has no closed-form algebraic expression. This is solved, essentially, for $x(t)$. We could use the error function, if you wish: $$ x(t) = \frac{\frac{1}{2}t\sqrt{\pi}\;\mathrm{erf}(t) + Ct}{e^{-t^2}}$$

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There are many solutions, of the form $te^{t^2} \{A + \int_{-\infty}^t e^{-s^2} \,ds\}$. Choosing the constant appropriately, by observing that $\lim_{t\to\infty} \int_{-\infty}^t e^{-s^2}\,ds$ is finite, fixes the desired solution and lets one compute the limit. –  Eugene Shvarts Feb 11 '13 at 8:04

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