If the solution of the differential equation 1+e^xydx+e^xy1-x/ydy=0 is x+kye^xy=C (where, C is an arbitrary constant), then the value of k is equal to

Question

MARKS Content Team · Accepted Answer

Correct answer: 1. The given equation is e^xydx-x/ydy+e^xydy+dx=0 or e^xyydx/y+e^xydy+dx=0 ⇒de^xyy+dx=0 On integrating, we get, e^xyy+x=C ⇒k=1

If the solution of the differential equation $(1 + e^{\frac{x}{y}}) d x + e^{\frac{x}{y}} (1 - \frac{x}{y}) d y = 0$ is $x + k y e^{\frac{x}{y}} = C$ (where, $C$ is an arbitrary constant), then the value of $k$ is equal to

Solution

More Differential Equations Questions

If the solution of the differential equation 1+exydx+exy1-xydy=0 is x+kyexy=C (where, C is an arbitrary constant), then the value of k is equal to

Solution

More Differential Equations Questions

If the solution of the differential equation $(1 + e^{\frac{x}{y}}) d x + e^{\frac{x}{y}} (1 - \frac{x}{y}) d y = 0$ is $x + k y e^{\frac{x}{y}} = C$ (where, $C$ is an arbitrary constant), then the value of $k$ is equal to