Question #312156

Solve



x^2y′′+ xy′− y =1/x + 1

1
Expert's answer
2022-03-16T12:52:18-0400

a0xnyn+a1xn1yn+...+an1xy+any=f(x)a_0x^ny^n+a_1x^{n-1}y^n+...+a_{n-1}xy'+a^ny=f(x)

x=eux=e^u

a0λ(λ1)(λ2)...(λn+1)...+an2λ(λ1)+an1λ+an=0a_0\lambda(\lambda-1)(\lambda-2)...(\lambda-n+1)...+a_{n-2}\lambda(\lambda-1)+a_{n-1}\lambda+a_n=0

(λ1)λ+λ1=0(\lambda-1)\lambda+\lambda-1=0

λ21=0\lambda^2-1=0

yy=1eu+1y''-y=\frac{1}{e^u}+1

λ1=1\lambda_1=1

λ2=1\lambda_2=-1

y=Ceu+C1euy=Ce^u+\frac{C_1}{e^u}


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.

Solution for 1:

a+bi=0, then s=0

y0=Ay_0=A

y0=0y''_0=0

A=-1

y0=1y_0=-1

Solution for 1eu\frac{1}{e^u}

a+bi=-1

s=1

y1=Aueuy_1=\frac{Au}{e^u}

y1=Au2Aeuy''_1=\frac{Au-2A}{e^u}

2Aeu=1eu\frac{-2A}{e^u}=\frac{1}{e^u}

A=-0.5

y1=u2euy_1=-\frac{u}{2e^u}

y=Ceuu2eu+C1eu1y=Ce^u-\frac{u}{2e^u}+\frac{C_1}{e^u}-1

u=ln(x)

y=ln(x)2x+Cx+C1x1y=\frac{-ln(x)}{2x}+Cx+\frac{C_1}{x}-1





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