Solve this nonhomegenous ode \(\displaystyle{y}{''}+{4}{y}={\cos{{\left({2}{x}\right)}}}\)

Nadia Clayton

Nadia Clayton

Answered question

2022-03-22

Solve this nonhomegenous ode
y+4y=cos(2x)

Answer & Explanation

Aarlenlsi1

Aarlenlsi1

Beginner2022-03-23Added 10 answers

Step 1
Then by variation of parameters we have the particular solution is given by
yp=cos(2x)u1+sin(2x)u2
where
u1=12cos(2x)sin(2x){rmd}x=116cos(4x)
and
u2=12cos(2x)cos(2x){rmd}x=x4+sin(4x)16.
Simplify, we have the general solution
y=c1cos(2x)+c2sin(2x)+14xsin(2x).
clarkchica44klt

clarkchica44klt

Beginner2022-03-24Added 17 answers

Step 1
Null space solution
m2+4=0
m=±2i
Hence, yN=C1sin2x+C2cos2x
Since, cos2x already in null space solution, take particular integral as
yp=x(Asin2x+Bcos2x)
Then,
yp{=Asin2x+Bcos2x+2x(Acos2xBsin2x)}
Then,
yp{=Asin2x+Bcos2x+2x(Acos2xBsin2x)}
yp=4Acos2x4Bsin2x4x(Asin2x+Bcos2x)
yp=4Acos2x4Bsin2x4x(Asin2x+Bcos2x)
yp+4yp=cos2x
4Acos2x4Bsin2x4x(Asin2x+Bcos2x)+4x(Asin2x+Bcos2x)=cos2x
4Acos2x4Bsin2x=cos2x
4A=1, 4B=0
A=14, B=0
Hence, yp=14xsin2x
So,
ycomplete=yN+yp
ycomplete=C1sin2x+C2cos2x+14{xsin2x}

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