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