Question #345991
  1. Find the general solution of the differential equation using the method

of variation of parameters y′′ + 4y = 4sec(2x)



1
Expert's answer
2022-05-30T16:25:33-0400

The corresponding homogeneous differential equation is


y+4y=0y''+4y=0

Corresponding (auxiliary) equation is


r2+4=0r^2+4=0

r1=2i,r2=2ir_1=2i, r_2=-2i

The general solution of the homogeneous differential equation is


yh=C1cos(2x)+C2sin(2x)y_h=C_1\cos(2x)+C_2\sin(2x)

Consider


y1=C1cos(2x)+C2sin(2x)y_1=C_1\cos(2x)+C_2\sin(2x)

y1=C1cos(2x)2C1sin(2x)+y_1'=C_1'\cos(2x)-2C_1\sin(2x)+

+C2sin(2x)+2C2cos(2x)+C_2'\sin(2x)+2C_2\cos(2x)

Let C1cos(2x)+C2sin(2x)=0.C_1'\cos(2x)+C_2'\sin(2x)=0. Then



y1=2C1sin(2x)+2C2cos(2x)y_1'=-2C_1\sin(2x)+2C_2\cos(2x)

y1=2C1sin(2x)4C1cos(2x)y_1''=-2C_1'\sin(2x)-4C_1\cos(2x)

+2C2cos(2x)4C2sin(2x)+2C_2'\cos(2x)-4C_2\sin(2x)

Substitute


2C1sin(2x)4C1cos(2x)-2C_1'\sin(2x)-4C_1\cos(2x)

+2C2cos(2x)4C2sin(2x)+2C_2'\cos(2x)-4C_2\sin(2x)




+4C1cos(2x)+4C2sin(2x)=4sec(2x)+4C_1\cos(2x)+4C_2\sin(2x)=4\sec (2x)

Solve the system


C1cos(2x)+C2sin(2x)=0C_1'\cos(2x)+C_2'\sin(2x)=02C1sin(2x)+2C2cos(2x)=4cos(2x)-2C_1'\sin(2x)+2C_2'\cos(2x)=\dfrac{4}{\cos(2x)}

C1cos(2x)+C2sin(2x)=0C_1'\cos(2x)+C_2'\sin(2x)=0C1sin(2x)cos(2x)+C2cos2(2x)=2-C_1'\sin(2x)\cos(2x)+C_2'\cos^2(2x)=2

C1cos(2x)+C2sin(2x)=0C_1'\cos(2x)+C_2'\sin(2x)=0C2=2C_2'=2


C1=2sin(2x)cos(2x)C_1'=-2\dfrac{\sin(2x)}{\cos(2x)}C2=2C_2'=2

Integrate


C1=2sin(2x)cos(2x)dx=lncos(2x)+C3C_1=-2\int \dfrac{\sin(2x)}{\cos(2x)}dx=\ln|\cos(2x)|+C_3

C2=2x+C4C_2=2x+C_4

The general solution of the non homogeneous differential equation is


y=lncos(2x)cos(2x)+C3cos(2x)y=\ln|\cos(2x)|\cos(2x)+C_3\cos(2x)+2xsin(2x)+C4sin(2x)+2x\sin(2x)+C_4\sin(2x)


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