Solve the Differential equations y'' + 4y = 4 tan^{2} x

Jason Farmer 2021-02-21 Answered
Solve the Differential equations y+4y=4tan2x
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Expert Answer

hajavaF
Answered 2021-02-22 Author has 90 answers
We need to solve following differential equation
y"+4y=4tan2x
Let us first solve the homogeneous part.
y"+4y=0
Let y=emx be the trial solution. Then we get the auxiliary equation
m2+4=0
This gives us
m=±2i
Thus the solution of the homogeneous part of the differential equation be
yh(x)=c1cos(2x)+c2sin(2x)
We use variation of parameter to calculate a
particular solution y_h. Let us take
y1=cos2x,y2=sin2x
Then we get
y1=2sin2x,y2=2cos2x
The wro
ian of y1 and y2 is
W=y1y2y1y2
=(cos2x)(2cos2x)(2sin2x)(sin2x)=20
In this case it is given that
f(x)=4tan(2x)
Then by the formula we obtain
u1=y2fWdx
=(sin2x)(4tan2x)2dx
=2sin2(2x)cos(2x)dx
=2(1cos22xcos2x)dx
=2(cos2xsec2x)dx
=2(cos2xsec2x)dx
=sin2xln|sec2x+tan2x|
Similarly
u2=y1fWdx
=(cos2x)(4tan2x)2dx
=2sin(2x)dx
=cos(2x)
Thus by variation of parameters the patricular solution is
yp=u1y1+u2y2
=(sin2xln|sec2x+tan2x|)(cos2x)+(cos2x)(sin2x)
yp=cos2xln|sec2x+tan2x|
Hence the general solution is
y=yc+yp=c1cos2x+c2sin2xcos2xln|sec2x+tan2x|
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