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Find the orthogonal trajectory of the curves 3x2y+2x3-y3-2y2 = ?, where &lpha is a constant
1 Answer
written 3.9 years ago by |
3x2y+2x2−y3−2y2=αDifferentiating this w.r.t x3(2xy+x2y′)+4x−3y2y′−4y y′=0To find the orthogonal trajectory, replacing y′by−1y′3(2xy−x2y′)+4x+3y2y′+4yy′=0
Multiplying throughout by y'3(2xyy′−x2)+4xy′+3y2+4y=0 6xyy′+4xy′−3x2+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 is∴3xy2−x3+4xy=αWhich is the orthogonal trajectory to3x2y+2x2−y3−2y2=α.