Integration By Substitution (1 Viewer)

[acevipa]

I'm having trouble with these last few questions from exercise 24e of fitzpatrick.

View reply for questions.

 

[acevipa]

1.

2.

 

[Drongoski]

dy/dr = 3/(1 - r)^4

Let: u = 1 - r Therefore du/dr = -1 or dr = -du

Therefore: y(r) = Int (3 x (1 - r)^(-4)) dr

= -3 Int u^(-4) du

= -3 (u ^(-4 + 1) / (-4 + 1) + C

= u^(-3) + C

= 1 / (1 - r)^3 + C

Given: y(0) = 1 / (1 - 0)^3 + C = 0

C = - 1

Therefore: y = 1 / (1 - r)^3 - 1

(1 - r)^3 = 1 / (y + 1)

1 - r = 1 / (y + 1)^(1/3)

Therefore: r = 1 - 1 / cuberoot (y + 1)



(Sorry: I'm hopeless at LaTeX ; so have to be long-winded above))