InterviewSolution
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InterviewSolution
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Solve y((dy)/(dx))^(2) |
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Answer» Solution :`y((dy)/(dx))^(2)+2x(dy)/(dx)-y=0` Solving quadratic in `(dy)/(dx)`, we GET `(dy)/(dx) = (-2x+-sqrt(4x^(2)+4y^(2)))/(2y)` `=(-X+_sqrt(x^(2)+y^(2)))/(y)` which is homogenous Put y=vx, i.e., `(dy)/(dx)=v+x(dv)/(dx)` Then given EQUATION transforms to `v+x(dv)/(dx)=(-1+-sqrt(1+v^(2)))/(v)` or `v^(2)+xv(dv)/(dx) -1+-sqrt(1+v^(2))` or `(v^(2)+1)+sqrt(1+v^(2))=-xv(dv)/(dx)` or `int(vdx)/((1+v^(2))+-sqrt(v^(2)+1))=-int(dx)/x` `or int(vdv)/(sqrt(v^(2)+1)(sqrt(1+v^(2)+-1)))= -int(dx)/(x)`................(1) Put `sqrt(1+v^(2))+-1` i.e., `v/sqrt(1+v^(2))dv=dt` Then, equation (1) transforms to `int(dt)/(t) = -int(dx)/(x)` or `"ln"t="ln"x+"ln"c` or `tx=c` or `(sqrt(1+v^(2))+-1)x=c` Given when `x=0, y=sqrt(5)` or `[sqrt(5)-0]=c` or `c=sqrt(5)` `THEREFORE x^(2)+y^(2)=5+x^(2)+-2sqrt(5)x` or `y^(2)=5+-2sqrt(5)x` |
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