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I have a problem in solving differential equation :

Let us consider following
y' - \frac{2}{t}y = t^2e^t,\qquad y(1)=0.

First as I understood, using definition of Lipschitz condition, this equation has unique solution; second using Euler's method algorithm for approximation this equation has following form

    set h=(b-a)/N;     t=a;     w=alpha;//( where alpha is initial value); output(t,w); for i=1,2.......N w=w+h*f(t,w)  (//f(t,y) is a function which we  should find) t=a+i*h; output(t,w) 

I know programming and can approximate it using program codes, but I want to solve it geometricaly or find unique solution, please help me.

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    just i forgot to write that y(1)=0,sorry for this2011-10-12

2 Answers 2

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Your equation is first order linear: y'+p(t)y=q(t) so it's solution is $ y = e^{-\int p(t)\,dt} \int e^{\int p(t)\,dt} q(t)\,dt$ See http://en.wikipedia.org/wiki/Linear_differential_equation subtopic "first order" for details.

You have $p(t) = -2/t$ so your integrating factor is $e^{\int -2/t \,dt} = e^{-2\ln(t)}=t^{-2}$. Thus $ y = t^2\int t^{-2}t^2e^t\,dt = t^2 \int e^t\,dt = t^2(e^t+C)$

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You can use an integrating factor.

The idea is to think that your left-hand side is actually the derivative of a product $\mu(t)y$, in which you have cancelled out a factor of $\mu(t)$. That is, you want to find a function $\mu(t)$ such that \mu(t)y' -\frac{2\mu(t)}{t}y = (\mu(t)y)'.

That means that you wan $\frac{d}{dt}\mu(t) = -\frac{2\mu(t)}{t}.$ This is separable, so you can solve it in the usual way: $\begin{align*} \frac{d\mu}{\mu} &= -\frac{2\,dt}{t}\\ \int\frac{d\mu}{\mu} &= -2\int\frac{dt}{t}\\ \ln|\mu| &= -2\ln |t| + C\\ |\mu| &= \frac{A}{t^2}\\ \mu(t) &= \frac{A}{t^2}.\end{align*}$

Selecting a simple one, say $\mu(t)=\frac{1}{t^2}$, leads to the expression in Bill Cook's answer: multiplying the equation through by $\frac{1}{t^2}$ we have: \begin{align*} y' - \frac{2}{t}y &= t^2e^t\\ \frac{1}{t^2}y' -\frac{2}{t^3}y &= e^t\\ \left(\frac{1}{t^2}y\right)' &= e^t\\ \int\left(\frac{1}{t^2}y\right)'\,dt &= \int e^t\,dt\\ \frac{1}{t^2}y &= e^t+C\\ y &= t^2(e^t + C) \end{align*}

Since you have $y(1)=0$, plugging in we get $0 = 1^2(e^1+C),$ so $C=-e$ and the solution is $y(t) = t^2(e^t - e).$

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    absolutely correct,exactly answer,thanks very much,thanks guys2011-10-12