initial value problem 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.
 A: 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
$$ y = t^2\int t^{-2}t^2e^t\,dt = t^2 \int e^t\,dt = t^2(e^t+C)$$
A: 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
$$y(t) = t^2(e^t - e).$$
