Author Topic: TUT 0701 Quiz 2  (Read 439 times)

maoyafei

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TUT 0701 Quiz 2
« on: October 07, 2019, 10:13:08 PM »
Question: (3(x^2)y+2*x*y+y^2) + (x^2+y^2)y' = 0
∂/∂y[𝜇(3(x^2)y+2*x*y+y^3)] = ∂/∂x[𝜇(x^2+y^2)]
∂𝜇/∂y (3(x^2)y+2*x*y+y^3) + ∂𝜇/∂x (x^2+y^2) + 𝜇(2*x)
∂𝜇/∂y (3(x^2)y+2*x*y+y^3) + 𝜇(3*x^2+3*y^2) = ∂𝜇/∂x (x^2+ y^2)
∂𝜇/∂y = 0      ∂𝜇/∂x = d𝜇/dx
𝜇(3*x^2+3*y^2) = d𝜇/dx (x^2+y^2)
(Divide both sides by (x^2+y^2)
3𝜇=d𝜇/dx
\int 1/𝜇d𝜇 = \int 3dx
ln(𝜇) = 3*x
𝜇=e^(3*x)