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Integration By Substitution (1 Viewer)
- Thread starter acevipa
- Start date
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))
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))
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