# Is this math correct for changing the variable of integration so I can put a function in it?

I have this differential equation $L/R*dy(t)/dt + y(t) = x(t)$ and the initial condition y(0) = 0

I use the integrating factor to get it into this form:

$y(t) *e^{\int R/L*dt} = \int x(t) *e^{\int R/L *dt} dt$

On the right hand side, can I change the variable of integration into some dummy variable, so I can stick the exponent inside of the integral?

$y(t) = 1/e^{R/L*t}*\int_0^t x(\tau)*e^{R/L*\tau} *d\tau = \int_0^t x(\tau) *e^{R/L(\tau-t)}*d\tau$

Why am I able to stick the t into the integrand, despite it not being a constant? Also, I'm confused on what's going on when I convert the indefinite integral into a definite integral and changing the variables of integration. Can someone explain this to me rigorously? Is $\int x(t)dt = \int_0^tx(\tau)d\tau$ always when x(0) = 0? What if the initial condition isn't 0?

## 1 Answer

$\int x(t)dt$ to my understanding it's a shorthand for $X(t)+c$ with $X(t)$ some primitive of $x(t)$, so is: $X'(t)=x(t)$. The expression $\int_0^tx(\tau)d\tau$ it's a more rigorous form to write the primitive since $\int_0^tx(\tau)d\tau=X(t)-X(0)$. So $c=-X(0)$

The initial condition can be set for $t=0$ or not, but anyway the initial condition sets the value of the constant $X(0)$

So, the rigorous way to write the primitive is $\int_0^tx(\tau)d\tau$, clearer when many variables are at play, as it happens in the general solution of your equation. The move you ask to clarify is $y(t) = 1/e^{R/L*t}*\int_0^t x(\tau)*e^{R/L*\tau} *d\tau = \int_0^t x(\tau) *e^{R/L(\tau-t)}*d\tau$ (it's not the right equation, but it's simpler and thus better to think about notation). $1/e^{R/L*t}$ it's constant from, so said, the point of view of the integrand in $\int_0^t x(\tau)*e^{R/L*\tau} *d\tau$, so you can put it under the integral symbol. Think of t being 32, do calculations with the number 32 in two places and then you move one 32 as you need to do, later 12, do calculations with the number 12 in two places... obviously, t in the upper limit of the integral is as constant as t in any other place.