Question #296532

𝑦′ − 𝑦 = 𝑒^2𝑥𝑦^3


1
Expert's answer
2022-02-14T17:22:54-0500

A Bernoulli differential equation, m=3,m0,m1m=3, m\not=0, m\not=1

Substitute v=y13=y2v=y^{1-3}=y^{-2}


v=2y3yv'=-2y^{-3}y'

yy=e2xy3y'-y=e^{2x}y^3

y3yy2=e2xy^{-3}y'-y^{-2} = e^{2x}

12vv=e2x-\dfrac{1}{2}v'-v=e^{2x}

v+2v=2e2xv'+2v=-2e^{2x}

Integrating factor


μ(x)=e(2)dx=e2x\mu(x)=e^{\int(2)dx}=e^{2x}

e2xv+2e2xv=2e4xe^{2x}v'+2e^{2x}v=-2e^{4x}

d(e2xv)=2e4xdxd(e^{2x}v)=-2e^{4x}dx

Integrate


d(e2xv)=2e4xdx\int d (e^{2x}v)=-\int2e^{4x}dx

e2xv=12e4x+Ce^{2x}v=-\dfrac{1}{2}e^{4x}+C

v=12e2x+12Ce2xv=-\dfrac{1}{2}e^{2x}+\dfrac{1}{2}Ce^{-2x}

Then


1y2=12e2x+12Ce2x\dfrac{1}{y^2}=-\dfrac{1}{2}e^{2x}+\dfrac{1}{2}Ce^{-2x}


y2=2Ce2xe2xy^2=\dfrac{2}{Ce^{-2x}-e^{2x}}

y2=2e2xCe4xy^2=\dfrac{2e^{2x}}{C-e^{4x}}


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