Question #126305
1. Solve the given differential equation using an appropriate substitution. The DE is a Bernoulli equation,

dy/dx = y(xy^6 - 1)

2. Solve the given differential equation by using an appropriate substitution. The DE is a Bernoulli equation.

x dy/dx + y = 1/y^2
1
Expert's answer
2020-07-20T18:00:59-0400
SolutionSolution

1. To solve the DE Bernoulli equation,

dy/dx=y(xy61)dy/dx = y(xy^6 - 1)


The standard form is

dydx+P(x)y=f(x)yn\frac{dy}{dx}+P(x)y=f(x)y^n


WhereWhere nn isis notnot equalequal toto zerozero oror oneone


dy/dx=xy7ydy/dx = xy^7 - y

dy/dx+y=xy7dy/dx+y = xy^7 >(1)-------------->(1)


n=7n=7

U=y1n=y17=y6U=y^{1-n}=y^{1-7}=y^{-6}

U=y6U=y^{-6}

U16=(y6)16U^{-\frac{1}{6}}=(y^{-6})^{-\frac{1}{6}}     \implies y=U16y=U^{-\frac{1}{6}}

dy/dx=16U76dudxdy/dx =- \frac{1}{6}U^{- \frac{7}{6}}*\frac{du}{dx}


16U76dudx+U16=xU76>(2)\frac{1}{6}U^{- \frac{7}{6}}\frac{du}{dx}+U^{-\frac{1}{6}}=x U^{- \frac{7}{6}}----------->(2)


Multiply by (16U76)( \frac{1}{6}U^{- \frac{7}{6}})


dudx6U1=6x\frac{du}{dx}-6U^1=-6x


y(x)=ϵρ(x)dx=ϵ6dx=ϵ6xy(x)=\epsilon^{\int \rho (x)dx}=\epsilon^{\int -6dx}= \epsilon^{-6x}

    y(x)=ϵ6x\implies y(x)=\epsilon^{-6x}


ϵ6xdudx6ϵ6xU=6xϵ6x\epsilon^{-6x} \frac{du}{dx}-6 \epsilon^{-6x} U=-6x\epsilon^{-6x}


ddx[ϵ6xU]=6xϵ6x\frac{d}{dx}[\epsilon^{-6x} U]=-6x\epsilon^{-6x}


Integration:

ϵ6xU=6xϵ6xdx\epsilon^{-6x} U=\int -6x\epsilon^{-6x} dx


ϵ6xU=xϵ6x+16ϵ6x+C\epsilon^{-6x} U=x\epsilon^{-6x} +\frac{1}{6}\epsilon^{-6x} +C


    U=x+16+Cϵ6x\implies U=x+\frac{1}{6}+\frac{C}{\epsilon^{-6x} }

y6=x+16+Cϵ6x>Answery^{-6}=x+\frac{1}{6}+C\epsilon^{6x} ------------->Answer




2. To solve the DE Bernoulli equation.


xdydx+y=1y2x \frac{dy}{dx} + y = \frac{1}{y^2}


The standard form is

dydx+P(x)y=f(x)yn\frac{dy}{dx}+P(x)y=f(x)y^n


Where n is not equal to zero or one


dydx+1xy=1xy2>(1)\frac{dy}{dx}+\frac{1}{x}y=\frac{1}{x}y^{-2}------------>(1)


n=2n=-2

U=y1n=y1(2)=y3U=y^{1-n}=y^{1-(-2)}=y^3

U=y3U=y^{3}

    y=U13\implies y=U^{\frac{1}{3}}


dydx=13U23dudx\frac{dy}{dx}=\frac{1}{3}U^{-\frac{2}{3}}\frac{du}{dx}


13U23dudx+1xU13=1xU23>(2)\frac{1}{3}U^{-\frac{2}{3}}\frac{du}{dx}+\frac{1}{x}U^{\frac{1}{3}}=\frac{1}{x}U^{-\frac{2}{3}}------->(2)


Multiply by (3U23)(3U^{-\frac{2}{3}})


dudx+3xU1=3x\frac{du}{dx}+\frac{3}{x}U^1=\frac{3}{x}


y(x)=ϵρ(x)dx=ϵ3xdx=ϵ3lnx=x3=x3y(x)=\epsilon^{\int \rho (x)dx}=\epsilon^{\int \frac{3}{x}dx}= \epsilon^{3ln|x|}=|x^3|=x^3

    y(x)=x3\implies y(x)=x^3


x3dudx+3x2U=3x2x^3\frac{du}{dx}+3x^2U=3x^2


ddx[x3U]=3x2\frac{d}{dx}[x^3U]=3x^2


x3U=3x23+Cx^3U=\frac{3x^2}{3}+C


U=1+Cx3U=1+\frac{C}{x^3}

y3=1+Cxx    y=(1+Cxx)13>Answery^3=1+\frac{C}{x^x} \implies y=(1+\frac{C}{x^x})^{\frac{1}{3}}------->Answer










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