The equation x^2ydx-(x3+y3)dy= 0 is homogeneous equation
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I think your question is: x² y dx - (x³ + y³) dy =0
That is x² y = (x³ + y³) (dy/dx)
(dy/dx) = x² y/ (x³ + y³) = 1/((x/y)+(y²/x²))
or, dx/dy = (x/y)+(y²/x²)
Integrating with respect to y, x= x log y + y³/(3 x²)
This on simplification will reduce to x³ = y³/3 (1 - log y)
That is x² y = (x³ + y³) (dy/dx)
(dy/dx) = x² y/ (x³ + y³) = 1/((x/y)+(y²/x²))
or, dx/dy = (x/y)+(y²/x²)
Integrating with respect to y, x= x log y + y³/(3 x²)
This on simplification will reduce to x³ = y³/3 (1 - log y)
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jaki23:
answer is right
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