Answer to Question #259850 in Differential Equations for Lalitha

Question #259850

Find the integral surface of the linear differential equation (x-y)y^2p+(y-x)x^2=(x^2+y^2)z, which passes through the curve xz=a^3,y=0

1
Expert's answer
2021-11-02T16:35:22-0400

(xy)y2p+(yx)x2q=(x2+y2)z(x-y)y^2p+(y-x)x^2q=(x^2+y^2)z


dx(xy)y2=dy(yx)x2=dz(x2+y2)z\frac{dx}{(x-y)y^2}=\frac{dy}{(y-x)x^2}=\frac{dz}{(x^2+y^2)z}


x2dx+y2dy=0x^2dx+y^2dy=0

x3+y3=c1x^3+y^3=c_1


dxdy(xy)(x2+y2)=dz(x2+y2)z\frac{dx-dy}{(x-y)(x^2+y^2)}=\frac{dz}{(x^2+y^2)z}


ln(xy)=lnz+lnc2ln(x-y)=lnz+lnc_2


xyz=c2\frac{x-y}{z}=c_2


F(c1,c2)=F(x3+y3,xyz)=0F(c_1,c_2)=F(x^3+y^3,\frac{x-y}{z})=0


for  the curve xz=a3,y=0xz=a^3,y=0 :


c1=x3c_1=x^3

x/z=c2x/z=c_2

    z=x/c2=c13/c2\implies z=x/c_2=\sqrt[3]{c_1}/c_2


x/c2=a3/x    x2=c2a3x/c_2=a^3/x\implies x^2=c_2a^3


z=zx3+y33xyz=\frac{z\sqrt[3]{x^3+y^3}}{x-y}


x3+y33=xy\sqrt[3]{x^3+y^3}=x-y


z=a3/x=a3/c13=a3x3+y33z=a^3/x=a^3/\sqrt[3]{c_1}=\frac{a^3}{\sqrt[3]{x^3+y^3}}


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