Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

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

share|cite|improve this question
P.S. You don't have an "initial value problem", since you are only giving a differential equation with no conditions on $y$. – Arturo Magidin Oct 12 '11 at 18:52
just i forgot to write that y(1)=0,sorry for this – dato datuashvili Oct 12 '11 at 18:57
up vote 4 down vote accepted

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 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 $$ y = t^2\int t^{-2}t^2e^t\,dt = t^2 \int e^t\,dt = t^2(e^t+C)$$

share|cite|improve this answer
thanks @Bill Cook – dato datuashvili Oct 12 '11 at 18:56

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 $$y(t) = t^2(e^t - e).$$

share|cite|improve this answer
absolutely correct,exactly answer,thanks very much,thanks guys – dato datuashvili Oct 12 '11 at 19:53

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.