ydy+(xy^2-8x)dx=0

differential equation
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1 Answer

ydy+(xy^2-8x)dx=0

ydy + x(y^2 - 8)dx = 0

y/(8 - y^2) dy = x dx

Integrating both sides,

int(y(8-y^2)) dy = int( x) dx

-(1/2).ln(8 - y^2) = x^2/2 + c

ln(8 - y^2) = A -- x^2

8 - y^2 = e^(A - x^2)

y^2 = 8 - e^(A - x^2)

by Level 11 User (81.5k points)

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