Author Topic: TUT 0202 Quiz  (Read 459 times)

Ruojing Chen

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TUT 0202 Quiz
« on: September 27, 2019, 02:11:52 PM »
dy/dx = ((x^2)+(3y^2))/(2xy)

dy/dt = (1+3y^2/x^2)/(2y/x)

let u = y/x, then y = ux

dy/dx = d(ux)/dx = (du/dx)*x + u = (1+3u^2) / (2u) = 1/(2u) + 3u/2
(du/dx)*x= 1/(2u) + u/2 = (1+u^2)/(2u)
∫ ( (2u) / (1+u^2) ) du = ∫ (1/x) dx
ln|1+(y/x)^2| = ln|x| + c

1+u = ax, a = e^c
1 + (y^2)/(x^2) - ax = 0
y^2 + x^2 - ax^3 = 0