Math, asked by krishnakowshikreddyv, 24 days ago

Solve
(y-xz)p + (yz-x)q = (x+y)(x-y)​

Answers

Answered by duraianna700
0

Answer:

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Answer to Question #110474 in Differential Equations for AMAN ADITYA

Answers>Math>Differential Equations

Question #110474

(y+xz)p-(x+yz)q=x²-y²

Expert's answer

Given, (y+xz)p-(x+yz)q = x^{2}-y^{2}(y+xz)p−(x+yz)q=x

2

−y

2

.

This equation is of the form Pp+Qq=RPp+Qq=R (Lagrange's linear partial differential equation).

Here, P = y+xz, Q=-(x+yz), R = x^{2}-y^{2}P=y+xz,Q=−(x+yz),R=x

2

−y

2

.

The subsidiary equations are

\dfrac{dx}{P}=\dfrac{dy}{Q}=\dfrac{dz}{R}\\ \dfrac{dx}{y+xz}=\dfrac{dy}{-(x+yz)}=\dfrac{dz}{x^{2}-y^{2}}~~~~~~~~~~~-(1)

P

dx

=

Q

dy

=

R

dz

y+xz

dx

=

−(x+yz)

dy

=

x

2

−y

2

dz

−(1).

Each ratio of (1) is equal to \dfrac{xdx+ydy}{(x^{2}-y^{2})z} = \dfrac{dx+dy}{(1-z)(y-x)}

(x

2

−y

2

)z

xdx+ydy

=

(1−z)(y−x)

dx+dy

.

Let us consider,

\dfrac{xdx+ydy}{(x^{2}-y^{2})z} = \dfrac{dz}{x^{2}-y^{2}}\\ xdx+ydy=zdz\\ \text{Integrating, }\\ \int xdx + \int ydy = \int zdz\\ \dfrac{x^{2}}{2}+\dfrac{y^{2}}{2} =\dfrac{z^{2}}{2}+\dfrac{c_{1}}{2}\\ x^{2}+y^{2}-z^{2}=c_{1}

(x

2

−y

2

)z

xdx+ydy

=

x

2

−y

2

dz

xdx+ydy=zdz

Integrating,

∫xdx+∫ydy=∫zdz

2

x

2

+

2

y

2

=

2

z

2

+

2

c

1

x

2

+y

2

−z

2

=c

1

Consider,

\dfrac{dx+dy}{(1-z)(y-x)}=\dfrac{dz}{(x+y)(x-y)}\\ \dfrac{dx+dy}{1-z}=\dfrac{dz}{-(x+y)}\\ -(x+y)d(x+y)=(1-z)dz\\ \text{Integrating,}\\ -\int (x+y)d(x+y)=\int (1+z)dz\\ -\dfrac{(x+y)^2}{2}=\dfrac{(1-z)^{2}}{-2}+\dfrac{c_{2}}{2}\\ (1-z)^{2}-(x+y)^{2}=c_{2}

(1−z)(y−x)

dx+dy

=

(x+y)(x−y)

dz

1−z

dx+dy

=

−(x+y)

dz

−(x+y)d(x+y)=(1−z)dz

Integrating,

−∫(x+y)d(x+y)=∫(1+z)dz

2

(x+y)

2

=

−2

(1−z)

2

+

2

c

2

(1−z)

2

−(x+y)

2

=c

2

The general solution is,

\phi(c_{1},c_{2})=0\\ \phi\left(x^{2}+y^{2}-z^{2}, (1-z)^{2}-(x+y)^{2}\right)=0.ϕ(c

1

,c

2

)=0

ϕ(x

2

+y

2

−z

2

,(1−z)

2

−(x+y)

2

)=0.

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