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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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  • $\begingroup$ P.S. You don't have an "initial value problem", since you are only giving a differential equation with no conditions on $y$. $\endgroup$ – Arturo Magidin Oct 12 '11 at 18:52
  • $\begingroup$ just i forgot to write that y(1)=0,sorry for this $\endgroup$ – dato datuashvili Oct 12 '11 at 18:57
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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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  • $\begingroup$ absolutely correct,exactly answer,thanks very much,thanks guys $\endgroup$ – dato datuashvili Oct 12 '11 at 19:53

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