Math, asked by arunkya, 2 months ago

z^2dx+(z^2-2yz)dy+(2y^2 solve​

Answers

Answered by farhaanaarif84
0

Answer:

Comparing the given equation with the standard equation Pdx+Qdy+Rdz=0,Pdx+Qdy+Rdz=0, we get

P=z^2,Q=z^2-2yz,R=2y^2-yz-xzP=z

2

,Q=z

2

−2yz,R=2y

2

−yz−xz

The given equation is homogeneous of degree 2. Now first of all, we test the condition of integrability.

P\left(\frac{\partial R}{\partial y}-\frac{\partial Q}{\partial z}\right)-Q\left(\frac{\partial R}{\partial x}-\frac{\partial P}{\partial z}\right)+R\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)\\ =z^2(4y-z-2z+2y)-(z^2-2yz)(-z-2z)+(2y^2-yz-xz)(0-0)\\ =6yz^2-3z^3+3z^3-6yz^2=0P(

∂y

∂R

∂z

∂Q

)−Q(

∂x

∂R

∂z

∂P

)+R(

∂x

∂Q

∂y

∂P

)

=z

2

(4y−z−2z+2y)−(z

2

−2yz)(−z−2z)+(2y

2

−yz−xz)(0−0)

=6yz

2

−3z

3

+3z

3

−6yz

2

=0

Hence, the equation is integrable.

Substitute x=uz,y=vzx=uz,y=vz into the equation.

This implies that, dx=udz+zdu,dy=vdz+zdvdx=udz+zdu,dy=vdz+zdv

z^2(udz+zdu)+z^2(1-2v)(vdz+zdv)+z^2(2v^2-v-u)dz=0z

2

(udz+zdu)+z

2

(1−2v)(vdz+zdv)+z

2

(2v

2

−v−u)dz=0

With this, the coefficient of dzdz will be 0. SO, we have;

z^3du+z^3(1-2v)dv=0z

3

du+z

3

(1−2v)dv=0

Divide through by z^3z

3

du+(1-2v)dv=0du+(1−2v)dv=0

Integrate through

u+v-v^2=Cu+v−v

2

=C

But, u=\frac{x}{z},y=\frac{y}{z}u=

z

x

,y=

z

y

\frac{x}{z}+\frac{y}{x}-\frac{y^2}{z^2}=C\\ xz+yz-y^2=Cz^2

z

x

+

x

y

z

2

y

2

=C

xz+yz−y

2

=Cz

2

Hence, the general solution of the equation is xz+yz-y^2=Cz^2xz+yz−y

2

=Cz

2

.

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