Question #299428

1. y²dx-(xy+y²)dy=0


Let substitution v=y/x ; y=vx ; dy= vdx+xdv



2. x²dy+(2xy-y²)dx=0


Let substitution v=y/x ; y=vx ; dy= vdx+xdv

1
Expert's answer
2022-02-21T05:54:22-0500

1. The given equation can be written as dydx=y2xy+y2\dfrac{dy}{dx} = \dfrac{y^2}{xy + y^2}

Put y=vx, then dydx=v+xdvdxy=vx, ~\text{then~} \dfrac{dy}{dx}= v + x\dfrac{dv}{dx} . The given equation becomes


v+xdvdx=v2x2vx2+v2x2=v2v+v2xdvdx=v1+v1=v21+v1+vv2dv=dxx1v2dv+1vdv=dxxIntegrating both sides, we get1v+logv=logx+logc1v=logvlogx+logc1v=(logv+logxlogc)1v=log(vxc)xy=log(yc)   (Using y=vx,v=yx)exy=ycy=cexy\begin{aligned} v + x\dfrac{dv}{dx} &= \dfrac{v^{2}x^2}{vx^2+v^2x^2}\\ &=\dfrac{v^2}{v+v^2}\\ x\dfrac{dv}{dx} &= \dfrac{v}{1+v} -1\\ &=-\dfrac{v^2}{1+v}\\ \dfrac{1+v}{v^2}dv & = -\dfrac{dx}{x}\\ \dfrac{1}{v^2}dv + \dfrac{1}{v}dv & = -\dfrac{dx}{x}\\ \end{aligned}\\ \text{Integrating both sides, we get}\\ \begin{aligned} -\dfrac{1}{v} + \log v &= -\log x + \log c\\ -\dfrac{1}{v} &= - \log v -\log x + \log c\\ -\dfrac{1}{v} &= - (\log v + \log x - \log c)\\ -\dfrac{1}{v} &= - \log\left(\frac{vx}{c}\right)\\ \dfrac{x}{y} &= \log\left(\frac{y}{c}\right)~~~(\text{Using $y = vx, v = \frac{y}{x}$})\\ e^{\frac{x}{y}} &= \frac{y}{c}\\ \therefore y &= ce^{\frac{x}{y}} \end{aligned}


2. The equation can be written as dydx=y22xyx2\dfrac{dy}{dx} = \dfrac{y^2 - 2xy}{x^2}

Put y=vx, then dydx=v+xdvdxy=vx, ~\text{then~} \dfrac{dy}{dx}= v + x\dfrac{dv}{dx} . The given equation becomes


v+xdvdx=v2x22vx2x2=v22vxdvdx=v23vdvv(v3)=dxxUsing partial fractions for the term on LHS, we get13[dvv3dvv]=dxxdvv3dvv=3dxxIntegrating both sides, we getlog(v3)logv=3logx+logclog(v3v)=logx3+logclog(v3v)=log(cx3)v3v=cx3   (taking anti-logarithms)13v=cx313xy=cx3   (Using y=vx)3xy=1cx3y=3x1cx3\begin{aligned} v + x\dfrac{dv}{dx} &= \dfrac{v^{2}x^2-2vx^2}{x^2}\\ &=v^2 - 2v\\ x\dfrac{dv}{dx} &= v^2 - 3v\\ \dfrac{dv}{v(v-3)} & = \dfrac{dx}{x}\\ \end{aligned}\\ \text{Using partial fractions for the term on LHS, we get}\\ \dfrac{1}{3}\left[\dfrac{dv}{v-3} - \dfrac{dv}{v}\right] = \dfrac{dx}{x}\\ \dfrac{dv}{v-3} - \dfrac{dv}{v} = 3\dfrac{dx}{x}\\ \text{Integrating both sides, we get}\\ \begin{aligned} \log (v-3) - \log v &= 3\log x + \log c\\ \log\left(\dfrac{v-3}{v}\right)&=\log x^3 + \log c\\ \log\left(\dfrac{v-3}{v}\right)&=\log (cx^3)\\ \dfrac{v-3}{v} &=cx^3~~~(\text{taking anti-logarithms})\\ 1 - \frac{3}{v} &=cx^3\\ 1 - \frac{3x}{y} &=cx^3~~~(\text{Using $y = vx$})\\ \frac{3x}{y} &= 1 - cx^3\\ \therefore y &= \frac{3x}{1 - cx^3} \end{aligned}

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