Question #281843

Solve the following differential equation by using the method of undetermined

coefficients:

š‘¦"+4š‘¦=3š‘„+š‘š‘œš‘ (2š‘„)


Expert's answer

characteristic equation:

k2+4=0k^2+4=0

k=±2ik=\pm2i

yh=c1cos2x+c2sin2xy_h=c_1cos2x+c_2sin2x


particular solution:

yp1=Ax+By_{p1}=Ax+B

A=3/4,B=0A=3/4,B=0


yp2=Axcos2x+Bxsin2xy_{p2}=Axcos2x+Bxsin2x

yp2′=Acos2x+Bsin2x+x(āˆ’2Asin2x+2Bcos2x)y'_{p2}=Acos2x+Bsin2x+x(-2Asin2x+2Bcos2x)

yp2′′=āˆ’2Asin2x+2Bcos2xāˆ’2Asin2x+2Bcos2x+x(āˆ’4Acos2xāˆ’4Bsin2x)y''_{p2}=-2Asin2x+2Bcos2x-2Asin2x+2Bcos2x+x(-4Acos2x-4Bsin2x)


āˆ’2Asin2x+2Bcos2xāˆ’2Asin2x+2Bcos2x+x(āˆ’4Acos2xāˆ’4Bsin2x)+-2Asin2x+2Bcos2x-2Asin2x+2Bcos2x+x(-4Acos2x-4Bsin2x)+

+4(Axcos2x+Bxsin2x)=cos2x+4(Axcos2x+Bxsin2x)=cos2x

āˆ’4A=0ā€…ā€ŠāŸ¹ā€…ā€ŠA=0-4A=0\implies A=0

4B=1ā€…ā€ŠāŸ¹ā€…ā€ŠB=1/44B=1\implies B=1/4


y=c1cos2x+c2sin2x+3x/4+xsin2x/4y=c_1cos2x+c_2sin2x+3x/4+xsin2x/4


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