Answer to Question #87047 in Differential Equations for Biraj

Question #87047
Solve: y2(x-y)dx+x2(x-y)dy+z(x2+y2)dz=0
1
Expert's answer
2019-03-26T12:15:45-0400
dxy2(xy)=dyx2(xy)=dxz(x2+y2){dx \over {y^2(x-y)}}={dy \over {x^2(x-y)}}=-{dx \over {z(x^2+y^2)}}

We chooze


P1=1z,Q1=1z,R1=xyP_1={1 \over z}, Q_1={1 \over z}, R_1=x-y

y2(xy)z+x2(xy)z(x2+y2)(xy)z=0{{y^2(x-y)} \over z}+{{x^2(x-y)} \over z}-{{(x^2+y^2)(x-y)} \over z}=0

Then function u1(x, y, z) is determined as


u1(x,y,z)=1zdx+1zdy+(xy)dxu_1(x, y, z)=\int{{1 \over z}dx}+\int{{1 \over z}dy}+\int{(x-y)dx}

xz+yz+(xy)z=C1{x \over z}+{y \over z}+(x-y)z=C_1

dxy2(xy)=dyx2(xy){dx \over {y^2(x-y)}}={dy \over {x^2(x-y)}}

dxy2=dyx2{dx \over {y^2}}={dy \over {x^2}}

y2dy=x2dxy^2dy=x^2dx

y3=x3+C2y^3=x^3+C_2

The equations


xz+yz+(xy)z=C1,y3=x3+C2{x \over z}+{y \over z}+(x-y)z=C_1, y^3=x^3+C_2

constitute the integral curves of the given differential equations.


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Comments

Assignment Expert
26.03.19, 19:18

Dear Biraj, You are welcome. We are glad to be helpful. If you liked our service please press a like-button beside the answer field. Thank you!

Biraj
26.03.19, 18:38

Thank you sir/ma'am I only reached upto y3=x3+c Thanks for another solution. Thank you so much.

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