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## Differential equations

### Course: Differential equations>Unit 1

Lesson 5: Exponential models

# Worked example: exponential solution to differential equation

AP.CALC:
FUN‑7 (EU)
,
FUN‑7.F (LO)
,
FUN‑7.F.1 (EK)
,
FUN‑7.F.2 (EK)
,
FUN‑7.G (LO)
,
FUN‑7.G.1 (EK)
The solution of the general differential equation dy/dx=ky (for some k) is C⋅eᵏˣ (for some C). See how this is derived and used for finding a particular solution to a differential equation.

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• Instead of putting the equation in exponential form, I differentiated each side of the equation:

(1/y) dy = 3 dx
ln y = 3x + C

Therefore

C = ln y - 3x

So, plugging in the given values of x = 1 and y = 2, I get that C = ln(2) - 3. If you put this in a calculator, it's a very different value (about -2.307) than what Sal got by raising both sides to the power of e: 2 * e^-3 evaluates to about .0996.

Why did my method not work? •  It's important to note that the C you work with is not the same as the C Sal works with. At , Sal gets e^C, which he says is just going to be another arbitrary constant. Thus Sal replaces e^c with C. If you perform e^(-2.307) you should get 0.0996.
• Sal changed e^C (which any arbitrary constant) to C (which is any another constant) and at least according to my knowledge C can't be any another constant as it will always be positive coz e^C can never be negative. We actually limited the value of our arbitrary constant to positive numbers, then shouldn't we mention that C is any constant number greater than zero i.e(C>0) ? • "Any arbitrary constant" includes complex numbers. Recall that e^(x*j) = cos(x) + j*sin(x). C could be, for example, 5 + π*j, in which case e^C ≈ -7.4.

However, even with complex exponents, e^C can never be zero! In this case that does not matter, because we have already limited ourselves to y ≠ 0 by using 1/y as part of the solution. But it is certainly something to keep in mind for other situations.
• At around , Sal decides to consider only the first scenario that he broke the absolute value function into, saying that "it will take the [other, negative part of it] into consideration". How did he know this would happen? • When dealing with separable equations why in some videos do you put y and the constant on one side and in others (like this one) you only move y over? • Wouldn't it be possible to solve for C right after you integrated both sides of the separated equation? ln |y| = 3x + C , when x = 1, y = 2.
Plug in these values and you'd get ln (2) = 3 + C, so C = ln (2) - 3
Then you'd get ln |y| = 3x +ln (2) - 3
So that y = e^(3x+ln (2) -3), because e^ln(2) = 2 you'd get 2e^(3x-3), and this is correct, but the method is different from the one Sal uses, can someone explain to me, why this works, or if there are any steps that aren't correct, so that you'd have to use Sal's method? • Why do we ultimately get the same specific solution when we solve for C with
y = Ce^(3x)
and substitute C back in and when we use
y = -Ce^(3x)
and substitute C back in?

I might be missing something fairly obvious, but since both functions satisfy our original differential equation and are the reflections of each other over the x-axis, there should be two functions that both pass through the point (1, 2) and satisfy dy/dx = 3y right?

Edit: I think (??) I answered my own question, but I would appreciate it if someone could verify it. In this case, since our two functions have C as a coefficient, and not as something that you add on, the two functions lead to the same specific solution. If you wanted to have the reflection over the x-axis of the solution also pass through the point, you would need to add a constant to it; but then, it wouldn't satisfy our differential equation. • If for whatever reason I had chosen to multiply both sides of the original integral in orange by -1, I would get very different solutions. I have always had a hard time understanding this. can someone help? • when I use y=-ce^3x then also I got the same ans y=e^(3x-3) as Sal sir got using y=ce^3x
that means no effect of absolute value here? • I solved the integral by using U-substitution (int(1/3y)dy=int(1)dx), but after evaluating, I got ln|3y|=3x+C instead of ln|y|=3x+C. What did I do wrong?

Here are my steps:

int(1/3y)dy=int(1)dx
int(1/3y)dy=x+C
(d/dy)(3y)=3
(1/3)int(3*1/3y)dy=x+C
u=3y
(1/3)int(1/u)du=x+C
(1/3)ln|u|+C=x+C
(1/3)ln|u|=x+C
ln|u|=3x+C
ln|3y|=3x+C
(1 vote) • when c=ln(3)/8 how do i set that to base e
(1 vote) 