Given: y′′+4y=cos2x
Require to solve the given differential equation.
Auxiliary equation corresponding to the given homogeneous differential equation y′′+4y=0
is m2+4=0
Solve the equation m2+4=0
Now m2+4=0⇒m=±2i
General solution to the homogeneous differential equation is
yh=c1cos2x+c2sin2x
Now let us find the particular solution.
Use the method of variation of parameters to find the particular solution.
Using the method of variation of parameters, Particular solution is yp=Au+Bv,=cos2x,v=sin2x and
A=−∫uv′−vu′vg(x)dx,B=∫uv′−vu′ug(x)dx , where g(x)=cos2x
First let us find uv′−vu′
Now u=cos2x,v=sin2x⇒uv′−vu′=(cos2x)(2cos2x)−(sin2x)(−2sin2x)=2cos22x+2sin22x=2(1)=2
Using the above, we get
A=−∫uv′−vu′vg(x)dx⇒A=−∫2(sin2x)(cos2x)dx=−∫2(2sinxcosx)(cos2x)dx
⇒A=−∫t3(−dt),t=cosx⇒dt=−sinxdx
Then, we get A=−4t4=−4cos4x
And
B=∫uv′−vu′ug(x)dx⇒B=∫2(cos2x)(cos2x)dx
⇒B=21∫(cos2x)[21+cos2x]dx
⇒B=41∫(cos2x)dx+41∫cos22xdx
⇒B=41∫(cos2x)dx+41∫21+cos4xdx
⇒B=41[2sin2x]+81[x+4sin4x]
⇒B=81sin2x+8x+32sin4x
Using the above, we have
yp=−41(cos4x)(cos2x)+[81sin2x+8x+32sin4x](sin2x)
Therefore general solution is
y=yh+yp
y=c1cos2x+c2sin2x−41(cos4x)(cos2x)+[81sin2x+8x+32sin4x](sin2x)
Comments