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Find the orthogonal trajectory of the curves 3x2y+2x3-y3-2y2 = ?, where &lpha is a constant
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3x2y+2x2y32y2=αDifferentiating this w.r.t x3(2xy+x2y)+4x3y2y4y y=0To find the orthogonal trajectory, replacing yby1y3(2xyx2y)+4x+3y2y+4yy=0

Multiplying throughout by y'3(2xyyx2)+4xy+3y2+4y=0 6xyy+4xy3x2+3y2+4y=0(6xy+4x) dydx 3x2+3y2+4y=0(6xy+4x) dy+(3x2+3y2+4y)dx=0

Hence the Differential equation is an exact equation. f(x, y)=M dx=(3x2+3y2+4y) dx= 3x33+3xy2+4xy+f(y)= x3+3xy2+4xy+f(y)

f(x,y)=Ndy=(6xy+4x)dy=6xy22+4xy+g(x)=3xy2+4xy+g(x)

Comparing the above two, f(y)=0, g(x)= x3So, the general solution is3xy2x3+4xy=αWhich is the orthogonal trajectory to3x2y+2x2y32y2=α.

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