Integral of 1/(1-x) is -ln(abs(1-x)). why negative?

[sparkle123]

Why is int(1/(1-x))=-ln(abs(1-x))?
If you substitute u=1-x
then
int(u)=ln(abs(u))=ln(abs(1-x))

Also, in this question you're trying to find the particular solution y=f(x) to the given differential euqation with initial conditions
f(-1)=2
dy/dx=6x^2-x^2y
d^2y/dx^2=-12.

So i get to:
ln(abs( 6 − y)) = − 1/3 x^3 − (1/3 − ln 4)
Why does the solution say I can remove absolute values to get:
y=6-4e^(-1/3(x^3+1)) ?

Thanks!

 

[SammyS]

If you let u = (1-x), then du = - dx .

 

[sparkle123]

Thanks SammyS!
Could someone please help me with the second part?

 

[Mark44]

Also, in this question you're trying to find the particular solution y=f(x) to the given differential euqation with initial conditions
f(-1)=2
dy/dx=6x^2-x^2y
d^2y/dx^2=-12.

I don't understand the 2nd and 3rd equations above. The first is your initial condition. Is the 2nd your differential equation? The 3rd doesn't follow from the 2nd.

Also, does x^2y mean (x²)y or x^2y?

So i get to:
ln(abs( 6 − y)) = − 1/3 x^3 − (1/3 − ln 4)
Why does the solution say I can remove absolute values to get:
y=6-4e^(-1/3(x^3+1)) ?

 

[sparkle123]

f(-1)=2
dy/dx=6x^2-x^2y
d^2y/dx^2=-12
are given conditions

I meant dy/dx=6x^2-(x^2)y

Essentially my question is why can you remove the absolute value signs.

Thanks! :D

 

[Mark44]

f(-1)=2
dy/dx=6x^2-x^2y
d^2y/dx^2=-12
are given conditions

I meant dy/dx=6x^2-(x^2)y

Essentially my question is why can you remove the absolute value signs.

Thanks! :D

What you have written about the 2nd derivative is very confusing, in part because it is incorrect.
The differential equation is dy/dx = 6x² - x²y
The initial conditions are y(-1) = 2 and y''(-1) = -12.

The second derivative is NOT -12, but the value of the second derivative at x = -1 is -12.

By separating variables, I was able to rewrite the diff. eqn. as dy/(6 - y) = x²dx. Integrating each side, I got -ln|6 - y| = (1/3)x³ + C, which is equivalent to
ln|6 - y| = (-1/3)x³ + C₂

From the first initial condition I found that C₂ = -1/3 + ln4, so
ln|6 - y| = (-1/3)x³ -1/3 + ln4 = (-1/3)(x³ + 1) + ln4

Exponentiation results in
|6 - y| = e^{-1/3(x^3 + 1) + ln4} = e^{ln4}e^{-1/3(x^3 + 1)}

Removing the absolute value signs, we have either
6 - y = e^{ln4}e^{-1/3(x^3 + 1)}

or
-(6 - y) = e^{ln4}e^{-1/3(x^3 + 1)}

Solving for y in each of these equations is relatively straightforward, and I leave that to you. One of these functions is such that y''(-1) = -12. The one for which this is true is your solution.

 

[sparkle123]

Oh! Thanks Mark44!