(px-y)(py+x)=2p clairuates equation
Answers
(px−y)(x−py)=2p
(pxy-y^2)(x-py)=2py(pxy−y
2
)(x−py)=2py
({py \over x}\cdot x^2-y^2)(1-{py \over x})=2{py \over x}(
x
py
⋅x
2
−y
2
)(1−
x
py
)=2
x
py
{py \over x}\cdot x^2-y^2=\dfrac{2\dfrac{py}{x}}{1-\dfrac{py}{x}}
x
py
⋅x
2
−y
2
=
1−
x
py
2
x
py
y^2={py \over x}\cdot x^2-\dfrac{2\dfrac{py}{x}}{1-\dfrac{py}{x}}y
2
=
x
py
⋅x
2
−
1−
x
py
2
x
py
Let x^2=u, y^2=v.x
2
=u,y
2
=v. Then
du=2xdx, dv=2ydydu=2xdx,dv=2ydy
p=\dfrac{dy}{dx}=\dfrac{2x}{2y}\cdot \dfrac{dv}{du}=\dfrac{\sqrt{u}}{\sqrt{v}}\cdot P, P=\dfrac{dv}{du}p=
dx
dy
=
2y
2x
⋅
du
dv
=
v
u
⋅P,P=
du
dv
{py \over x}=\dfrac{dv}{du}=P
x
py
=
du
dv
=P
Substitute
v=Pu-\dfrac{2P}{1-P}v=Pu−
1−P
2P
We have Clairaut Equation
v=Pu+f(P)v=Pu+f(P)
The general solution is given by v=Cu+f(C), C v=Cu+f(C),C is an arbitrary constant.
Therefore
v=Cu-\dfrac{2C}{1-C}, C \ \text{is arbitrary constant}, C\not=1v=Cu−
1−C
2C
,C is arbitrary constant,C
=1
The general solution is
y^2=Cx^2-\dfrac{2C}{1-C}, \ C \ \text{is arbitrary constant}, C\not=1y
2
=Cx
2
−
1−C
2C
, C is arbitrary constant,C
=1
Answer:
Solve the equation
(px-y)(py+x)= 2p put X=x^2 Y =y^2