Here is problem I am trying to solve for my differential equations class. I have spent several hours trying to solve it and have not been successful. I have tried breaking up the fraction various ways and using different techniques of integration but can't seem to get a solution that is valid near $t=\frac{1}{2}$. Here is the problem;

Solve the IVP $$\frac{dy}{dt}= t + \frac{t}{t^2−1}y ; \quad y\left(\frac{1}{2}\right) = 6.$$ Hint: Be careful with your integrating factor as you need to describe a solution that is valid near $t = 1/2.$

I am trying to solve the problem in the form: $$\frac{dy}{dt} - \frac{t}{t^2−1}y = t $$ using an integrating factor as the hint suggests.

If anyone could suggest some further hints as to how to solve this problem I would appreciate it. I would like to figure it out without having the solution and steps given to me.

  • 2
    $\begingroup$ If you follow the procedure for finding the integrating factor correctly, you'll eventually get to the DE $$\frac{d}{dt}\Bigg(\frac{y}{\sqrt{1-t^2}}\Bigg)=\frac{t}{\sqrt{1-t^2}}$$ $\endgroup$
    – Matthew H.
    Feb 18, 2021 at 4:57
  • 1
    $\begingroup$ @MatthewPilling And this leads to the correct solution $$y=t^2-1+\frac{9\sqrt{3}}{2} \sqrt{1- t^2}$$ $\endgroup$
    – Raffaele
    Feb 18, 2021 at 11:30
  • $\begingroup$ Thank you, I was able to figure it out using this information. I hadn't thought of bringing the negative sign into the denominator. $\endgroup$
    – DoctorDave
    Feb 18, 2021 at 17:46

1 Answer 1


$$\frac{dy}{dt}= t + \frac{t}{t^2−1}y$$ In the present case the method of integrating factor isn't the simplest. Nevertheless we will use it as requested.

You wrote : I am trying to solve the problem in the form $\frac{dy}{dt}- \frac{t}{t^2−1}y=t$ using an integrating factor.

This is a bad start because the correct form to start is $$N(t,y)dt+M(t,y)dy=0$$ You should start with this form : $$\left(t + \frac{t}{t^2−1}y\right)dt-dy=0$$ Then one have to find an integrating factor $\mu(t,y)$ so that the equation $$\mu(t,y)\left(t + \frac{t}{t^2−1}y\right)dt-\mu(t,y)dy=0\quad\text{be exact}.$$


Try the simplest cases of function $\mu(t,y)$. For example try $\mu$ function of $t$ only. You will find $$\mu=\frac{1}{\sqrt{t^2-1}}$$

  • $\begingroup$ $\mu$ is not real at $t=1/2$. $\endgroup$
    – Raffaele
    Feb 18, 2021 at 11:20
  • $\begingroup$ This doesn't matter. The integrating factor allows to find the general solution of the ODE without considering any condition. Once the general solution is found it is time to consider the condition and see if a particular solution exists or not and is real or not. $\endgroup$
    – JJacquelin
    Feb 18, 2021 at 18:19

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