What is the general solution of the differential equation? : # dy/dx + (2x)/(x^2+1)y=1/(x^2+1) #

3 Answers
Aug 16, 2017

# y = x/(x^2+1) + C/(x^2+1) #

Explanation:

We have:

# dy/dx + (2x)/(x^2+1)y=1/(x^2+1) # ..... [A]

This is a first order linear 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 \ (2x)/(x^2+1) ) \ dx ) #
# \ \ = exp( ln|x^2+1| ) #
# \ \ = exp( ln(x^2+1) ) # as #x^2+1 gt 0#
# \ \ = x^2+1 #

And if we multiply the DE [A] by this Integrating Factor, #I#, we will (by virtue of the IF) have a perfect product differential;

# (x^2+1)dy/dx + (2x)y=1 #

# :. d/dx((x^2+1)y) = 1 #

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

# (x^2+1)y = int \ dx #

And integrating gives us:

# (x^2+1)y = x + C #

Which we can rearrange to get:

# y = x/(x^2+1) + C/(x^2+1) #

Aug 16, 2017

#y(x) = (x+ C)/(x^2 + 1)#

Explanation:

GIven: #dy/dx + (2x)/(x^2 + 1)y = 1/(x^2 + 1)#

I am going to make a slight change of notation and mark it as equation [1]:

#y'(x) + (2x)/(x^2 + 1)y(x) = 1/(x^2 + 1)" [1]"#

Equation [1] is in the form:

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

where #P(x) = (2x)/(x^2+1)# and #Q(x)=1/(x^2+1)#

This type of equation is known to have an integrating factor:

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

Substitute for #P(x)#

#mu(x) = e^(int(2x)/(x^2+1)dx#

Integrate:

#mu(x) = e^(ln(x^2+1))#

Simplify:

#mu(x) = x^2+1#

Multiply both sides of the equation [1] by #mu(x)#:

#(x^2 + 1)y'(x) + (2x)y(x) = 1#

Set up both sides for integration:

#int((x^2 + 1)y'(x) + (2x)y(x))dx = intdx#

We do not actually integrate the left side integrates, because multiplication by the integrating factor assures that the left side integrates to #mu(x)y(x)#. We do integrate right side but the integral is trivial:

#(x^2 + 1)y(x) = x + C#

Solve for #y(x)#:

#y(x) = (x+ C)/(x^2 + 1)#

Aug 16, 2017

#y = (x+C_3)/(x^2+1)#

Explanation:

This differential equation is completely equivalent to

#(x^2+1)(dy)/(dx)+2x y = 1# because #x^2+1 > 0 forall x in RR#

The homogeneous part of this linear differential equation is

#(x^2+1)y'_h+2x y_y = 0# which is separable

#(2x)/(x^2+1) dx = -dy_h/(y_h)# and integrating

#log_e(x^2+1) = -log_e y_h + C_1# then

#x^2+1=C_2/y_h# or

#y_h = C_2/(x^2+1)#

Substituting now #y_p = (C_2(x))/(x^2+1)# into the complete equation we obtain

#C_2'(x) = 1# and then

#C_2(x) = x + C_3#

Now the complete solution is obtained as #y = y_h+y_p# and then

#y = (x+C_3)/(x^2+1)#