Answer to Question #200732 in Differential Equations for Gaurav Kumar

Question #200732

(π‘₯ βˆ’ 𝑦)𝑦2𝑝 + (𝑦 βˆ’ π‘₯)π‘₯2π‘ž = (π‘₯2 + 𝑦2)𝑧,

Find the Integra surface when π‘₯𝑧 = π‘Ž3 (π‘Ž > 0), 𝑦 = 0.



1
Expert's answer
2021-05-31T16:39:21-0400

dxxβˆ’y)y2=dy(yβˆ’x)x2=dz(x2+y2)z\frac{dx}{x-y)y^2}=\frac{dy}{(y-x)x^2}=\frac{dz}{(x^2+y^2)z}

dxxβˆ’y)y2=dy(yβˆ’x)x2\frac{dx}{x-y)y^2}=\frac{dy}{(y-x)x^2}

x2dx=βˆ’y2dyx^2dx=-y^2dy

x3+y3=C1x^3+y^3=C_1

dxβˆ’dyxy2βˆ’y3βˆ’yx2+x3=dz(x2+y2)z\frac{dx-dy}{xy^2-y^3-yx^2+x^3}=\frac{dz}{(x^2+y^2)z}

d(xβˆ’y)(xβˆ’y)(x2+y2)=dz(x2+y2)z\frac{d(x-y)}{(x-y)(x^2+y^2)}=\frac{dz}{(x^2+y^2)z}

d(xβˆ’y)xβˆ’y=dzz\frac{d(x-y)}{x-y}=\frac{dz}{z}

ln(xβˆ’y)=ln(z)+ln(C2)ln(x-y)=ln(z)+ln(C_2)

C2=xβˆ’yzC_2=\frac{x-y}{z}

F(x3+y3,xβˆ’yz)=0F(x^3+y^3,\frac{x-y}{z})=0

For xz=a3, y=0

x3=C1x^3=C_1

x2=a3C2x^2=a^3C_2

The integral surface 

x=C


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