What is the general solution of the differential equation? : #y''+4y=2sin2x#

Question

Isaiah Allen · Accepted Answer

The general solution is #y=c_1cos(2x)+c_2sin(2x)-1/2xcos2x#

#y''+4y=2sin(2x)#

This ODE is non-homogenous, second order linear.

One way to express the general solution is as

#y=y_h+y_p#

Find #y_h# by solving

#y''+4y=0#

Complete the caracteristic equation.

#r^2+4=0#

#r=+-2i# where #i^2=-1#

The resolution is

#y_h=c_1cos(2x)+c_2sin(2x)#

Locate a specific form solution.

#y_p=a_0xsin2x+a_1xcos2x#

#y_p'=2a_0xcos2x+a_0sin2x-2a_1xsin2x+a_1cos2x#

#y_p''=-4a_0xsin2x+2a_0cos2x+2a_0cos2x-4a_1xcos2x-2a_1sin2x-2a_1sin2x#

#=-4a_0xsin2x+4a_0cos2x-4a_1xcos2x-4a_1sin2x#

Filling in the ODE with those values

#y''+4y=2sin(2x)#

#-4a_0xsin2x+4a_0cos2x-4a_1xcos2x-4a_1sin2x+4(a_0xsin2x+a_1xcos2x)=2sin2x#

#-4a_0xsin2x+4a_0cos2x-4a_1xcos2x-4a_1sin2x+4a_0xsin2x+4a_1xcos2x=2sin2x#

#4a_0cos2x-4a_1sin2x=2sin2x#

#<=>#, #{(a_0=0),(a_1=-1/2):}#

#y_p=-1/2xcos2x#

The general solution is #y=c_1cos(2x)+c_2sin(2x)-1/2xcos2x#

Evelyn Agard · Answer

# y(x) = Acos(2x)+Bsin(2x) -1/2xcos(2x) #

We have: # y'' + 4y=2sin2x# ..... [A] This is a second order non-Homogeneous Differentiation Equation. The standard approach is to find a solution, #y_c# of the homogeneous equation by looking at the Auxiliary Equation, which is the quadratic equation with the coefficients of the derivatives, and then finding an independent particular solution, #y_p# of the non-homogeneous equation. Complementary Role The equation homogeneous linked to [A] is # y''+4y= 0 # Additionally, the related auxiliary equation is: # m^2+4 = 0 # Which has two pure imaginary solutions #m = +-2i# Consequently, the homogeneous equation's solution is: # y_c = e^(0x){Acos(2x)+Bsin(2x)} # # \ \ \ = Acos(2x)+Bsin(2x) # Specific Resolution For this specific equation [A], the following is a likely solution: # y = acos(2x)+bsin(2x) # Where #a# and #b# are constants to be determined by substitution. However, this solution is part of #y_c#, thus we seek an alternate independant solution of the form: # y = axcos(2x)+bxsin(2x) # Let us assume the above solution works, in which case be differentiating wrt #x# we have: # y' \ \= (b-2ax)sin2x+(2bx+a)cos2x# # y'' = -4(ax-b)cos2x-4(bx+a)sin2x # Substituting into the initial Differential Equation #[A]# we get: # {-4(ax-b)cos2x-4(bx+a)sin2x} + 4{axcos(2x)+bxsin(2x)} = 2sin2x # Equating coefficients of #cos(2x)# and #sin(2x)# we get: #cos(2x): 4b = 0 => b =0# #sin(2x): -4a=2 => a=-1/2 # Thus, we arrive at the specific solution: # y_p = -1/2xcos(2x) # Overall Resolution which ultimately results in the GS of [A} # y(x) = y_c + y_p # # \ \ \ \ \ \ \ = Acos(2x)+Bsin(2x) -1/2xcos(2x) #

Axel Alvarez · Answer

undefined The general solution of the given differential equation $ y'' + 4y = 2\sin(2x) $ is $ y(x) = c_1 \cos(2x) + c_2 \sin(2x) + \frac{1}{2} \sin(2x) $, where $ c_1 $ and $ c_2 $ are arbitrary constants.