# Thread: Is this a general "formula" for IVP problems?

1. ## Is this a general "formula" for IVP problems?

y(t) = [ 1/u(t) ][ int(u(t)g(t)ds) + y_o ]

I'm currently learning IVP's right now. I noticed for some of the more simpler IVPs I just plugged in y=1 and t=0 for say y(0)=1 to find C and the particular solution.

But I came across this "formula" in my notes and I noticed that my professor used it for two different examples.

I'm curious as to whether this works for any differential equation or did it happen to only be those two examples?

(and sorry, my ability to use latex is poor)

2. ## Re: Is this a general "formula" for IVP problems?

As written, I don't think that formula is good for much of anything, as there is no s-dependence of the integrand. That is, the RHS of your equation there simplifies down much more than, I think, is intended.

It looks a bit like the formula for a linear first-order solution. Do your notes have this equation listed in connection with a DE?

3. ## Re: Is this a general "formula" for IVP problems?

That formula also is not useful because you didn't tell us what "u" and "g" mean!

It is true that if the differential equation a(t)y'+ b(t)y= g(t), a linear equation then there exist an "integrating factor", u(t) such such that a(t)u(t)y'+ b(t)u(t)= d(u(t)y(t))/dt, then multiplying by u(t) makes the equation d(u(t)y(t))/dt= g(t)u(t) which is equivalent to d(u(t)y(t))= g(t)u(t)dt and integrating gives $u(t)y(t)= \int g(t)u(t)dt$ which we can convert to a definite integral that gives $y(t_0)= y_0$ by $u(t)y(t)= \int_{t_0}^t u(s)g(s)ds+ y_0$

Dividing by u(t), $y(t)= \frac{1}{u(t)}\int_{t_0}^t u(x)g(s)ds$.

That is a general solution for first order, linear differential equations. It does not apply to higher order differential equations and it does not apply to non-linear differential equations.