# Linear Differential Equation Using Integrating Factor

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.

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

$$\frac{dy}{dt}= t + \frac{t}{t^2−1}y$$ 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 $$\mu=\frac{1}{\sqrt{t^2-1}}$$
• $\mu$ is not real at $t=1/2$. – Raffaele Feb 18 at 11:20