Alternate method of solving a differential equation.

Today I encountered an alternate way of solving differential equations. Let's say that we have the following differential equation:

$$\frac{dP}{dt} = 0.9P$$

Step $1$: Divide both sides of equation by $P$ and this will yield

$$\frac{1}{P} \ \frac{dP}{dt} = 0.9$$

Step $2$: Replace $\frac{1}{P} \ \frac{dP}{dt}$ with $[ln(P)]'$ and this yields

$$[ln(P)]' = 0.9$$

Step $3$: We write integrals with respect to $t$ on both sides and we have

$$\int [ln(P)]'dt = \int 0.9 dt$$

Step $4$: Use $FTOC$ to integrate and we get

$$ln(P) = 0.9t + C$$

I already know that we get as our general solution $P = Ce^{0.9t}$, but why not just do separation of variables? Wouldn't that just make the work much easier than doing it this way? I would also like some clarification on Step $2$, especially where we had to replace $\frac{1}{P} \ \frac{dP}{dt}$ with $[ln(P)]'$... I can't quite grasp that just yet.

• the dervivative of ln f is $\frac 1 f \frac {df}{dx}$ or write $\frac {d \ln(f)} {df} \frac {df}{dx}=\frac 1 f \frac {df}{dx}$ see chain rule Sep 27, 2017 at 0:59

Step 2: We claim that: $\frac{1}{P} \frac{dP}{dt} = [\ln(P)]'$.

This is actually easier to prove going from right to left. Just use the chain rule and differentiate the right hand side. Let $h(x) = [\ln(x)]'$. Then

$[\ln(P)]' = h(P) * P' = \frac{1}{P} * P' = \frac{1}{P} \frac{dP}{dt}$.

No one is forcing you to do it one way or the other, but from a mathematical point of view this way is more rigorous, as it does not involve moving infinitesimals around and treats the Leibniz notation as a symbol for the derivative.

• Thanks for the response. It's just that I'm used to treating $\frac{dP}{dt}$ as a "ratio" and therefore can do separation of variables. Aside from that, your explanation did clear my concerns. Sep 27, 2017 at 1:06
• Yeah, in "standard analysis" i.e. rigorous calculus we treat the fraction dP/dt as a symbol to mean the derivative, however there is also "non-standard analysis" in which the use of the fraction as a ratio is rigorously treated.
– Ð..
Sep 27, 2017 at 1:14
• Ah now that makes sense now. It's basically a matter of how we look at $\frac{dP}{dt}$ then. Sep 27, 2017 at 1:16

For this one, there probably isn't any reason to not do separation of variables. But the technique used illustrates a different way to do it, which is similar to a method used for solving first order linear differential equations.

Showing this method, could be a stepping stone for introducing that method.

• Hi videlity. I was actually introduced to this method today, but we ran our of time and my professor rushed this part, therefore making some of the students confused, including myself. But after studying it and doing a few practice problems, I can get used to this method. Sep 27, 2017 at 1:08