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) #
