# What is the particular solution of the differential equation # y' + y tanx = sin(2x) # where #y(0)=1#?

##### 2 Answers

#### Explanation:

The given equation,

, is of the form:

Where

It is known that the integrating factor is:

Multiply the given equation by

We know that the left side integrates to

Multiply both side by

Use the boundary condition to find the value of C:

# y = -2cos^2x + 3cosx #

#### Explanation:

We have

# y' + y tanx = sin(2x) #

Which can be written:

# y' + tanx \ y = sin(2x) #

This is a first order linear ordinary differential equation of the form:

# (d zeta)/dx + P(x) zeta = Q(x) #

We solve this using an Integrating Factor

# I = exp( \ int \ P(x) \ dx ) #

# \ \ = exp( int \ (tanx) \ dx ) #

# \ \ = exp( ln |secx| ) #

# \ \ = exp( ln secx ) #

# \ \ = secx #

And if we multiply the DE by this Integrating Factor,

# secx y' + y secxtanx = secx sin(2x) #

# :. d/dx(ysecx) = secx sin(2x) #

Which is now a separable DE, so we can "separate the variables" to get:

# ysec x = int \ secx sin(2x) \ dx #

# " " = int \ 1/cosx (2sinxcosx) \ dx #

# " " = 2 \ int \ sinx \ dx #

And, Integrating we get:

# ysecx = -2cosx + C #

# :. y/cosx = -2cosx + C #

# :. y = -2cos^2x + Ccosx #

Which is the general Solution. Then using

# 1 = -2cos^2 0 + Ccos0 => 1 = -2 + C => C= 3#

Hence

# :. y = -2cos^2x + 3cosx #