Question #312155

Solve

x ^2y'′ + 2xy'− 12y = x^3log x.



1
Expert's answer
2022-03-16T14:33:12-0400

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

x=eu

(λ−1)λ+2λ−12=0(\lambda-1)\lambda+2\lambda-12=0

λ2+λ−12=0\lambda^2+\lambda-12=0

Îť1=3\lambda_1=3

λ2=−4\lambda_2=-4

y′′+y′−12y=ue3uy''+y'-12y=ue^{3u}

y0=Ce3u+C1e4uy_0=Ce^{3u}+\frac{C_1}{e^{4u}}

Solution for ue3uue^{3u}

a+bi=3, then s=1

y1=u(Au+B)e3uy_1=u(Au+B)e^{3u}

y1′=(3Au2+(3B+2A)u+B)e3uy'_1=(3Au^2+(3B+2A)u+B)e^{3u}

y1′′=(9Au2+(9B+12A)u+6B+2A)e3uy''_1=(9Au^2+(9B+12A)u+6B+2A)e^{3u}

14Aue3u+(7B+2A)e3u=ue3u14Aue^{3u}+(7B+2A)e^{3u}=ue^{3u}


A=1/14

B=-1/49

y1=(u14−149)ue3uy_1=(\frac{u}{14}-\frac{1}{49})ue^{3u}

y=(u14−149)ue3u+Ce3u+C1e4uy=(\frac{u}{14}-\frac{1}{49})ue^{3u}+Ce^{3u}+\frac{C_1}{e^4u}




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