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

Question

Charles Arocha · Accepted Answer

#y = -2cos^2(x)+3cos(x)#

The given equation,

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

, is of the form:

#y' + P(x)y = Q(x)#

Where #P(x) = tan(x), and Q(x) = sin(2x)#

It is known that the integrating factor is:

#mu(x) = e^(intP(x)dx)#

#inttan(x)dx = log(sec(x))#

#mu(x) = e^log(sec(x)) = sec(x)#

Multiply the given equation by #mu(x)#:

#y'sec(x)+tan(x)sec(x)y=sin(2x)/sec(x)#

We know that the left side integrates to #mu(x)y# and we are left with the task of integrating the right side:

#sec(x)y = intsin(2x)sec(x)dx#

#sec(x)y = -2cos(x)+C#

Multiply both side by #cos(x)#

#y = -2cos^2(x)+Ccos(x)#

Use the boundary condition to find the value of C:

#1 = -2cos^2(0)+Ccos(0)#

#C = 3#

Evelyn Agard · Answer

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

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, #I#, we will have a perfect product differential; # 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 #y(0)=1# we can find #C# as: # 1 = -2cos^2 0 + Ccos0 => 1 = -2 + C => C= 3# Hence # :. y = -2cos^2x + 3cosx #

Ezra Adams · Answer

undefined The particular solution of the given differential equation $ y' + y \tan(x) = \sin(2x) $ with the initial condition $ y(0) = 1 $ is $ y(x) = \sin(x) + \cos(x) $.