q) x² - 2 (x + y) - y2
Answers
Answer:
Here
P=y^2(x-y), Q=x^2(x-y),Q=z(x^2+y^2)P=y
2
(x−y),Q=x
2
(x−y),Q=z(x
2
+y
2
)
{dx \over y^2(x-y)}={dy \over x^2(x-y)}={dz \over z(x^2+y^2)}=
y
2
(x−y)
dx
=
x
2
(x−y)
dy
=
z(x
2
+y
2
)
dz
=
From 1st and 2nd fraction, we get
(x^2(x-y))dx=(y^2(x-y))dy(x
2
(x−y))dx=(y
2
(x−y))dy
Dividing by (x-y),(x−y), we get
x^2dx-y^2dy=0x
2
dx−y
2
dy=0
d(x^3-y^3)=0d(x
3
−y
3
)=0
Integrating, we get
x^3-y^3=c_1x
3
−y
3
=c
1
If we take
P_1={1 \over y}, Q_1=-{1 \over x},R_1={1 \over z}P
1
=
y
1
,Q
1
=−
x
1
,R
1
=
z
1
then
PP_1+QQ_1+RR_1=PP
1
1
+RR
1
=
={1 \over y}(y^2(x-y))-{1 \over x}(x^2(x-y))+{1 \over z}(z(x^2+y^2))==
y
1
(y
2
(x−y))−
x
1
(x
2
(x−y))+
z
1
(z(x
2
+y
2
))=
=xy-y^2-x^2+xy+x^2+y^2=0=xy−y
2
−x
2
+xy+x
2
+y
2
=0
Thus for the given system of equations, we have
{dx \over y^2(x-y)}={dy \over x^2(x-y)}={dz \over z(x^2+y^2)}=
y
2
(x−y)
dx
=
x
2
(x−y)
dy
=
z(x
2
+y
2
)
dz
=
={dx/y-dy/x+dz/z \over 0}=
0
dx/y−dy/x+dz/z
{dx \over y}=-{dy \over x}={dz \over z}
y
dx
=−
x
dy
=
z
dz
From 1st and 2nd fraction, we get
{dx \over y}=-{dy \over x}
y
dx
=−
x
dy
xdx+ydy=0xdx+ydy=0
x^2+y^2=c_2x
2
+y
2
=c
2
Let F be an arbitrary differentiable function. Then the general equation of a partial differential equation is
F(x^3-y^3,x^2+y^2)=0F(x
3
−y
3
,x
2
+y
2
)=0
Step-by-step explanation: