Negative Natural Log when Solving dy/dx = 100 - y ; with y(0) = 60; for x=1 and x=2

[koreamaniac101]

The equation i am asked to solve is this:

dy/dx = 100 - y ; with y(0) = 60; for x = 1 and x = 2

Seems simple enough, but when they present the solution:

dy/(100 - y) = dx => -ln|100 - y| = x + C => 100 - y = Ae^-x => Using the initial condition, we find that A = 40.

y(1) = 100 - 40e^-1
y(2) = 100 - 40e^-2


My confusion lies in the negative ln. When you integrate 1/(100 - y), shouldnt it just be a regular positive ln(100 - y)?

Furthermore, why do they have a negative x exponenet on the other side?

How I solved it (incorrectly) is shown below:

dy/(100 - y) = dx

Integrate both sides:

ln|100 - y| = x + C

Plug in x = 0, and y = 60:

C = 40

ln|100 - y| = x + ln40

Bring everything to the power of e:

100 - y = e^{x + ln40}

100 - y = 40e^x

y = 100 - 40e^x



Can someone show me what i did wrong?

 

[Deleted member 4993]

The equation i am asked to solve is this:

dy/dx = 100 - y ; with y(0) = 60; for x = 1 and x = 2

Seems simple enough, but when they present the solution:

dy/(100 - y) = dx => -ln|100 - y| = x + C => 100 - y = Ae^-x => Using the initial condition, we find that A = 40.

y(1) = 100 - 40e^-1
y(2) = 100 - 40e^-2


My confusion lies in the negative ln. When you integrate 1/(100 - y), shouldnt it just be a regular positive ln(100 - y)?

Furthermore, why do they have a negative x exponenet on the other side?

How I solved it (incorrectly) is shown below:

dy/(100 - y) = dx

Integrate both sides:

ln|100 - y| = x + C

Plug in x = 0, and y = 60:

C = 40

ln|100 - y| = x + ln40

Bring everything to the power of e:

100 - y = e^{x + ln40}

100 - y = 40e^x

y = 100 - 40e^x



Can someone show me what i did wrong?

If you differentiate

ln|100 - y|

What do you get?

If you differentiate

ln|100 + y|

What do you get?

Do you see the difference?

 

[JeffM]

\(\displaystyle p < 0 \implies \displaystyle \int \dfrac{dp}{p } = -\ \int \dfrac{dp}{-\ p} = -\ \{ ln(-\ p) * (-\ 1) - C \} = ln(|p|) + C.\)

\(\displaystyle p > 0 \implies \displaystyle \int \dfrac{dp}{p } = ln(p) + C = ln(|p|) + C.\)

The general form is \(\displaystyle \displaystyle \int \dfrac{dp}{p } = ln(|p|) + C.\)

 

[HallsofIvy]

Equivalent to what the others have written: let u= 100- y. Then du= -dy or dy= -du.

\(\displaystyle \int \frac{dy}{100- y}= \int \frac{-du}{u}= -\int \frac{du}{u}= -ln(u)+ C= -ln(100- y)+ C\).