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

Question

Charles Anctil · Accepted Answer

The General Solution is:

# y = -1/2x -1/4 + Ce^(2x) #

We can use an integrating factor when we have a First Order Linear non-homogeneous Ordinary Differential Equation of the form; # dy/dx + P(x)y=Q(x) # We have: # dy/dx = x+2y # Which we can write as: # dy/dx -2y = x # ..... [A] This is a First Order Ordinary Differential Equation in Standard Form. So we compute and integrating factor, #I#, using; # I = e^(int P(x) dx) # # \ \ = exp(int \ -2 \ dx) # # \ \ = exp( -2x) # # \ \ = e^(-2x) # And if we multiply the DE [A] by this Integrating Factor, #I#, we will have a perfect product differential; # dy/dxe^(-2x) -2ye^(-2x) = xe^(-2x) # # :. d/dx( ye^(-2x) ) = xe^(-2x) # This has transformed our initial ODE into a Separable ODE, so we can now "separate the variables" to get:: # ye^(-2x) = int \ xe^(-2x) \ dx # We can proceed via an application of integration by Parts Let # { (u,=x, => (du)/dx,=1), ((dv)/dx,=e^(-2x), => v,=-1/2e^(-2x) ) :}# Then plugging into the IBP formula: # int \ (u)((dv)/dx) \ dx = (u)(v) - int \ (v)((du)/dx) \ dx # We have: # int \ (x)(e^(-2x)) \ dx = (x)(-1/2e^(-2x)) - int \ (-1/2e^(-2x))(1) \ dx # # :. int \ xe^(-2x) \ dx = -1/2xe^(-2x) +1/2 \ int \ e^(-2x) \ dx # # " " = -1/2xe^(-2x) -1/4 e^(-2x) + C # Using this result, we can write the DE Solution as : # ye^(-2x) = -1/2xe^(-2x) -1/4 e^(-2x) + C # # :. y = -1/2x -1/4 + Ce^(2x) #

William Abdalla · Answer

undefined The general solution to the differential equation dy/dx = x + 2y is y = Ce^(2x) - 0.5x - 0.5, where C is an arbitrary constant.

Isabella Ackerman · Answer

undefined The general solution of the given differential equation dy/dx = x + 2y is y(x) = Ce^(2x) - (1/2)x - (1/4), where C is an arbitrary constant.