# Non-exact 1st Order Differential Equation? Kindly solve this Non-exact 1st Order Differential Equation by integrating Factor Method. #(x+2)sinydx+x cosydy=0#

##### 1 Answer

Jan 8, 2018

# y = arcsin(e^(C-x)/x^2) #

#### Explanation:

We have:

# (x+2)siny \ dx + xcosy \ dy = 0 #

Which we can write in standard form as:

# xcosy \ dy/dx + (x+2)siny = 0 #

We note that we do not require the use of an Integrating Factor as this is a separable First Order differential equation, thus we can collect terms and separate the variables to get

# \ \ \ \ \ cosy/siny \ dy/dx + (x+2)/x = 0 #

# :. int \ coty \ dy = - int \ (x+2)/x \ dx #

And we integrate to get:

# ln siny = -x - 2lnx + C#

Exponentiating we get:

# siny = e^(-x - 2lnx + C)#

# \ \ \ \ \ \ \ = e^(C-x)e^(- 2lnx) #

# \ \ \ \ \ \ \ = e^(C-x)e^(ln(1/x^2)) #

# \ \ \ \ \ \ \ = e^(C-x)(1/x^2) #

# \ \ \ \ \ \ \ = e^(C-x)/x^2 #

And finally:

# y = arcsin(e^(C-x)/x^2) #