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

Question

Emma Aldridge · Accepted Answer

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

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) #

Emilia Aguilar · Answer

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

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)#

Emily Ammerman · Answer

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

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)#

Ezekiel Alexander · Answer

undefined The general solution of the given differential equation $ \frac{{dy}}{{dx}} + \frac{{2x}}{{x^2+1}}y = \frac{{1}}{{x^2+1}} $ is:

$$ y = \frac{{e^{-\int \frac{{2x}}{{x^2+1}} \, dx}}}{{x^2+1}} \left( \int \frac{{e^{\int \frac{{2x}}{{x^2+1}} \, dx}}}{{x^2+1}} \, dx + C \right) $$

where $ C $ is the constant of integration.