Answer to Question #253999 in Differential Equations for Fah

Question

write the ordinary differential equation (1+sin y)dx = (2ycosy-x(secy-tan y))dy

Answer 1

(1+sin y)dx = (2ycosy-x(secy+tan y))dy

=> $\frac{dx}{dy} +x \frac{secy+tany}{1+siny}= \frac{2ycosy}{1+siny}$

=> $\frac{dx}{dy} +x \frac{(1+siny)}{cosy(1+siny)}= \frac{2ycosy}{1+siny}$

=> $\frac{dx}{dy} +x secy= \frac{2ycosy}{1+siny}$

This is a linear differential equation

Integrating factor (I.F.) = $e^{\int (secy )dy}= e^{ln(secy +tany)}$

= secy + tany = $\frac{1+siny}{cosy}$

Multiplying both sides by I.F. we get

$\frac{1+siny}{cosy}.\frac{dx}{dy} +x \frac{(secy+tany}{1+siny}.\frac{1+siny}{cosy}= \frac{2ycosy}{1+siny}.\frac{1+siny}{cosy}$

=> d($x\frac{1+siny}{cosy}$ )= 2ydy

Integrating we get

$\int d( x\frac{1+siny}{cosy})= \int2ydy+C$

=> $x\frac{1+siny}{cosy}= y² + C$ , where $C$ is integration constant.

This is the solution of the differential equation given.