The corresponding homogeneous differential equation is
y′′+4y=0 Corresponding (auxiliary) equation is
r2+4=0
r1=2i,r2=−2i The general solution of the homogeneous differential equation is
yh=C1cos(2x)+C2sin(2x) Consider
y1=C1cos(2x)+C2sin(2x)
y1′=C1′cos(2x)−2C1sin(2x)+
+C2′sin(2x)+2C2cos(2x)
Let C1′cos(2x)+C2′sin(2x)=0. Then
y1′=−2C1sin(2x)+2C2cos(2x)
y1′′=−2C1′sin(2x)−4C1cos(2x)
+2C2′cos(2x)−4C2sin(2x) Substitute
−2C1′sin(2x)−4C1cos(2x)
+2C2′cos(2x)−4C2sin(2x)
+4C1cos(2x)+4C2sin(2x)=4sec(2x) Solve the system
C1′cos(2x)+C2′sin(2x)=0−2C1′sin(2x)+2C2′cos(2x)=cos(2x)4
C1′cos(2x)+C2′sin(2x)=0−C1′sin(2x)cos(2x)+C2′cos2(2x)=2
C1′cos(2x)+C2′sin(2x)=0C2′=2
C1′=−2cos(2x)sin(2x)C2′=2Integrate
C1=−2∫cos(2x)sin(2x)dx=ln∣cos(2x)∣+C3
C2=2x+C4
The general solution of the non homogeneous differential equation is
y=ln∣cos(2x)∣cos(2x)+C3cos(2x)+2xsin(2x)+C4sin(2x)
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