Question #312153

8. Solve x

x^2y′′− 2xy′+ 2y = x^4sin(4 log x)


1
Expert's answer
2022-03-17T06:56:34-0400

x=eux=e^u

a0λ(λ−1)(λ−2)...(λ−n+1)...an−2λ(λ−1)+an−1λ+an=0a_0\lambda(\lambda-1)(\lambda-2)...(\lambda-n+1)...a_{n-2}\lambda(\lambda-1)+a_{n-1}\lambda+a_n=0

(λ−1)λ−2λ+2=0(\lambda-1)\lambda-2\lambda+2=0

λ2−3λ+2=0\lambda^2-3\lambda+2=0

λ1=2\lambda_1=2

λ2=1\lambda_2=1

y=Ceu+C1e2uy=Ce^u+C_1e^{2u}

yi=useau(Rm(u)cos(bu)+Tm(u)sin(bu))y_i=u^se^{au}(R_m(u)cos(bu)+T_m(u)sin(bu))

s=0 if a+bi is not a root or s=k if it is.

a+bi=4i+4 s=0

y0=e4u(Bsin(4u)+Acos(4u))y_0=e^{4u}(Bsin(4u)+Acos(4u))

y0′(u)=(4B−4A)e4usin(4u)+(4B+4A)e4ucos(4u)y'_0(u)=(4B-4A)e^{4u}sin(4u)+(4B+4A)e^{4u}cos(4u)

y0′′=32Be4ucos(4u)−32Ae4usin(4u)y''_0=32Be^{4u}cos(4u)-32Ae^{4u}sin(4u)

(−10B−20A)e4usin(4u)+(20B−10A)e4ucos(4u)=e4usin(4u)(-10B-20A)e^{4u}sin(4u)+(20B-10A)e^{4u}cos(4u)=e^{4u}sin(4u)

A=−125A=\frac{-1}{25}

B=−150B=\frac{-1}{50}

y0=e4u(−sin(4u)50−cos(4u)25)y_0=e^{4u}(\frac{-sin(4u)}{50}-\frac{cos(4u)}{25})

y=e4u(−sin(4u)50−cos(4u)25)+Ceu+C1e2uy=e^{4u}(\frac{-sin(4u)}{50}-\frac{cos(4u)}{25})+Ce^u+C_1e^{2u}






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