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The solution of the differential equ `(x^2 + y^2) dx = 2 xy dy` is- |
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Answer» Given, `(x^(2)+y^(2)) dx - 2xy dy = 0 ` or ` (x^(2)+y^(2)) dx = 2xy dy` or ` (dy)/(dx) = (x^(2)+y^(2))/(2xy) ` ....(i) Let y = vx Thus `(dy)/(dx) = v + x (dv)/(dx)` Thus ` v+ x (dv)/(dx) = (x^(2)+(vx)^(2))/(2x(vx))` or ` v + x (dv)/(dx) = (1+ v^(2))/(2v)` or ` x (dv)/(dx) = (1+v^(2))/(2v) - v` or ` x (dv)/(dx) = (1+v^(2)-2v^(2))/(2v)` or ` x (dv)/(dx) = (1-v^(2))/(2v)` or ` (dx)/x = (2v)/(1-v^(2)) dv` or ` (dx)/x - (2v)/(1-v^(2)) dv = 0` ...(ii) Integrating both sides, ` log x + log (1-v^(2)) = log C` or ` log x (1 - v^(2)) = log C` or ` x (1 - v^(2)) = C` or ` x (1 - y^(2)/x^(2)) = C` or ` x ((x^(2)-y^(2))/x^(2)) = C` or ` x^(2) - y^(2) = Cx` |
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