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junior college 2 | H2 Maths
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Date Posted: 3 years ago
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You can't differentiate the right side with respect to x, as it's not in terms of x.
dy/dx = y(1 - y)
1/(y(1 - y)) dy/dx = 1
Next , either use partial fractions method to split it up or do the following :
(1 - y + y)/(y(1 - y)) dy/dx = 1
(1/y + 1/(1 - y) ) dy/dx = 1
1/y dy/dx + 1/(1 - y) dy/dx = 1
Differentiate both sides with respect to x,
∫ (1/y dy/dx + 1/(1 - y) dy/dx) dx = ∫ 1 dx
ln|y|- ln|1 - y| = x + c, c is a constant
ln|y/(1 - y)|= x + c
|y/(1 - y)| = eˣ+ᶜ
y/(1 - y) = eˣ+ᶜ or y/(1 - y) = -eˣ+ᶜ
y = (1 - y)(eˣ+ᶜ) or y = (y - 1)(eˣ+ᶜ)
y + yeˣ+ᶜ = eˣ+ᶜ or yeˣ+ᶜ - y = eˣ+ᶜ
y(1 + eˣ+ᶜ) = eˣ+ᶜ or y(eˣ+ᶜ - 1) = eˣ+ᶜ
y = eˣ+ᶜ / (1 + eˣ+ᶜ) or y = eˣ+ᶜ / (eˣ+ᶜ - 1)
y = eˣ / (1/eᶜ + eˣ) or y = eˣ / (eˣ - 1/eᶜ)
y = eˣ / (A + eˣ) or y = eˣ / (eˣ - A), where A = 1/eᶜ
1/(y(1 - y)) dy/dx = 1
Next , either use partial fractions method to split it up or do the following :
(1 - y + y)/(y(1 - y)) dy/dx = 1
(1/y + 1/(1 - y) ) dy/dx = 1
1/y dy/dx + 1/(1 - y) dy/dx = 1
Differentiate both sides with respect to x,
∫ (1/y dy/dx + 1/(1 - y) dy/dx) dx = ∫ 1 dx
ln|y|- ln|1 - y| = x + c, c is a constant
ln|y/(1 - y)|= x + c
|y/(1 - y)| = eˣ+ᶜ
y/(1 - y) = eˣ+ᶜ or y/(1 - y) = -eˣ+ᶜ
y = (1 - y)(eˣ+ᶜ) or y = (y - 1)(eˣ+ᶜ)
y + yeˣ+ᶜ = eˣ+ᶜ or yeˣ+ᶜ - y = eˣ+ᶜ
y(1 + eˣ+ᶜ) = eˣ+ᶜ or y(eˣ+ᶜ - 1) = eˣ+ᶜ
y = eˣ+ᶜ / (1 + eˣ+ᶜ) or y = eˣ+ᶜ / (eˣ+ᶜ - 1)
y = eˣ / (1/eᶜ + eˣ) or y = eˣ / (eˣ - 1/eᶜ)
y = eˣ / (A + eˣ) or y = eˣ / (eˣ - A), where A = 1/eᶜ
Thank you so much, J.
No prob
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