1.

Solve: \(\frac {dy}{dx} \) = 1 - xy + y - x.

Answer»

\(\frac {dy}{dx} \) = 1-xy+y-x

= 1-x + y (1-x)

= (1-x) (1+y)

∴  \(\frac {dy}{1+y} \) = (1-x) dx

= ln [1+y] = \(x-\frac {x^2}{2} + C\) ( By integrating both sides)

which is solution of given differential equation.



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