2

I am currently learning about ode's and I am on the topic of linear non-autonomous ode's. The professor is deriving the solution to the following ode using some integrating factor $I$.

$$\frac{dx}{dt} = A(t)x+B(t)$$

So these are the steps that he makes in finding $x$.

$$\begin{align} I\left(\frac{dx}{dt} - Ax = B \right) \\ I\frac{dx}{dt} - AIx = IB \\ I\frac{dx}{dt} - \frac{dI}{dt} x = IB \\ d(Ix) = IB \\ \frac{dI}{dt} = -AI \end{align}$$

I understand all of those steps and now, since this a simple ode we can separate it and arrive at solution for $I$ as

$$I = \exp (\int -Adt) = e^{-\int A}$$

Now, takes this solution of $I$ and uses it in $d(Ix) = IB$ by integrating both sides and subbing in $I$.

$$\begin{align} Ix &= \int IB \\ x &= \frac{1}{I} \int IB \\ x &= e^{\int A} (\int B e^{-\int A}) & \text{?!} \\ \end{align}$$

That last bit is where my confusion is. How did he arrive at the last step because if we sub in $I$, should it not be $\frac{1}{e^{-\int A}}$? Where did the negative go and why is it reciprocal?

Thanks!

Jeel Shah
  • 9,306

2 Answers2

3

If I were he, I would write $I=e^{-\int A}+C$ instead, but if we're assumed to have $C=0$ then $I=e^{-\int A}$ is the correct one. Now, if $Ix=\int IB$, with the substitution $I=e^{-\int A}$, we certainly get his last solution. Note that $\frac{1}I=\frac{1}{e^{-\int A}}=e^{\int A}$.

Mikasa
  • 67,374
1

The confusion arises from that fact that I was thinking that $\frac{1}{e^{-\int A}}$ was different from $\frac{1}{e^{-x}}$ but that is not the case. The negative sign in front of the $x$ is what reciprocates it into $e^{\int A}$.

Jeel Shah
  • 9,306