BMTC-132 Solved Assignment 2025
DIFFERENTIAL EQUATIONS
a) Solve the differential equations:
d
y
d
x
=
−
2
x
+
y
+
1
4
x
+
2
y
−
1
d
y
d
x
=
−
2
x
+
y
+
1
4
x
+
2
y
−
1
(dy)/(dx)=-(2x+y+1)/(4x+2y-1) \frac{d y}{d x}=-\frac{2 x+y+1}{4 x+2 y-1} d y d x = − 2 x + y + 1 4 x + 2 y − 1 b) Solve the simultaneous equations:
d
x
y
−
z
=
d
y
z
−
x
=
d
z
x
−
y
d
x
y
−
z
=
d
y
z
−
x
=
d
z
x
−
y
(dx)/(y-z)=(dy)/(z-x)=(dz)/(x-y) \frac{d x}{y-z}=\frac{d y}{z-x}=\frac{d z}{x-y} d x y − z = d y z − x = d z x − y
a) By using the method of variation of parameter, find the general solution of the differential equation:
y
′
′
+
y
=
sec
2
x
y
′
′
+
y
=
sec
2
x
y^(”)+y=sec^(2)x y^{\prime \prime}+y=\sec ^2 x y ′ ′ + y = sec 2 x b) If :
Z
=
x
2
y
+
2
x
y
4
Z
=
x
2
y
+
2
x
y
4
Z=x^(2)y+2xy^(4) Z=x^2 y+2 x y^4 Z = x 2 y + 2 x y 4 where
x
=
sin
zt
x
=
sin
zt
x=sin zt \mathrm{x}=\sin \mathrm{zt} x = sin zt
and
y
=
cos
t
y
=
cos
t
quad y=cos t \quad y=\cos t y = cos t
find
dz
dt
dz
dt
(dz)/(dt) \frac{\mathrm{dz}}{\mathrm{dt}} dz dt when
t
=
0
t
=
0
t=0 \mathrm{t}=0 t = 0 by (i) chain rule and by (ii) the direct substitution.
a) Verify that the differential equation:
z
(
x
2
−
y
2
−
z
2
)
d
x
+
(
x
+
z
)
x
z
d
y
+
x
(
z
2
−
x
2
−
x
y
)
d
z
=
0
z
x
2
−
y
2
−
z
2
d
x
+
(
x
+
z
)
x
z
d
y
+
x
z
2
−
x
2
−
x
y
d
z
=
0
{:[z(x^(2)-y^(2)-z^(2))dx+(x+z)xzdy],[+x(z^(2)-x^(2)-xy)dz=0]:} \begin{aligned}
& z\left(x^2-y^2-z^2\right) d x+(x+z) x z d y \\
& +x\left(z^2-x^2-x y\right) d z=0
\end{aligned} z ( x 2 − y 2 − z 2 ) d x + ( x + z ) x z d y + x ( z 2 − x 2 − x y ) d z = 0 is integrable and find its integral.
y
′
=
z
y
4
+
x
4
x
y
3
y
′
=
z
y
4
+
x
4
x
y
3
y^(‘)=(zy^(4)+x^(4))/(xy^(3)) y^{\prime}=\frac{z y^4+x^4}{x y^3} y ′ = z y 4 + x 4 x y 3
a) Find the general solution of the equation:
x
(
y
2
−
z
2
)
u
x
+
y
(
z
2
−
x
2
)
u
y
+
z
(
x
2
−
y
2
)
u
z
=
0
x
y
2
−
z
2
u
x
+
y
z
2
−
x
2
u
y
+
z
x
2
−
y
2
u
z
=
0
x(y^(2)-z^(2))u_(x)+y(z^(2)-x^(2))u_(y)+z(x^(2)-y^(2))u_(z)=0 x\left(y^2-z^2\right) u_x+y\left(z^2-x^2\right) u_y+z\left(x^2-y^2\right) u_z=0 x ( y 2 − z 2 ) u x + y ( z 2 − x 2 ) u y + z ( x 2 − y 2 ) u z = 0 b) Show that for the function
f
f
f f f given by:
f
(
x
,
y
)
=
x
y
x
2
+
y
2
lim
x
→
0
lim
y
→
0
f
(
x
,
y
)
=
lim
y
→
0
lim
x
→
0
f
(
x
,
y
)
=
0
f
(
x
,
y
)
=
x
y
x
2
+
y
2
lim
x
→
0
lim
y
→
0
f
(
x
,
y
)
=
lim
y
→
0
lim
x
→
0
f
(
x
,
y
)
=
0
{:[f(x”,”y)=(xy)/(x^(2)+y^(2))],[lim_(x rarr0)lim_(y rarr0)f(x”,”y)=lim_(y rarr0)lim_(x rarr0)f(x”,”y)=0]:} \begin{gathered}
f(x, y)=\frac{x y}{x^2+y^2} \\
\lim _{x \rightarrow 0} \lim _{y \rightarrow 0} f(x, y)=\lim _{y \rightarrow 0} \lim _{x \rightarrow 0} f(x, y)=0
\end{gathered} f ( x , y ) = x y x 2 + y 2 lim x → 0 lim y → 0 f ( x , y ) = lim y → 0 lim x → 0 f ( x , y ) = 0 But
lim
(
x
,
y
)
f
(
x
,
y
)
lim
(
x
,
y
)
f
(
x
,
y
)
lim_((x,y))f(x,y) \lim _{(x, y)} f(x, y) lim ( x , y ) f ( x , y ) does not exist.
5. a) Show that the following function
f
f
f f f is differentiable at
(
0
,
0
)
(
0
,
0
)
(0,0) (0,0) ( 0 , 0 ) :
f
(
x
,
y
)
=
{
x
2
y
2
x
2
+
y
2
,
if
(
x
,
y
)
≠
(
0
,
0
)
0
,
if
(
x
,
y
)
=
(
0
,
0
)
f
(
x
,
y
)
=
x
2
y
2
x
2
+
y
2
,
if
(
x
,
y
)
≠
(
0
,
0
)
0
,
if
(
x
,
y
)
=
(
0
,
0
)
f(x,y)={[(x^(2)y^(2))/(x^(2)+y^(2))”,”,” if “(x”,”y)!=(0″,”0)],[0,”,”” if “(x”,”y)=(0″,”0)]:} f(x, y)=\left\{\begin{array}{cc}
\frac{x^2 y^2}{x^2+y^2}, & \text { if }(x, y) \neq(0,0) \\
0 & , \text { if }(x, y)=(0,0)
\end{array}\right. f ( x , y ) = { x 2 y 2 x 2 + y 2 , if ( x , y ) ≠ ( 0 , 0 ) 0 , if ( x , y ) = ( 0 , 0 ) b) Find the complete solution of the equation
p
x
+
q
y
=
p
q
p
x
+
q
y
=
p
q
px+qy=pq p x+q y=p q p x + q y = p q .
6. a) Using Charpit’s method, find the complete integral of the differential equation:
p
2
x
+
q
2
y
=
z
p
2
x
+
q
2
y
=
z
p^(2)x+q^(2)y=z p^2 x+q^2 y=z p 2 x + q 2 y = z b) Find partial derivatives
f
x
f
x
f_(x) f_x f x and
f
y
f
y
f_(y) f_y f y for the function:
f
(
x
,
y
)
=
5
x
4
y
2
+
6
x
2
y
3
f
(
x
,
y
)
=
5
x
4
y
2
+
6
x
2
y
3
f(x,y)=5x^(4)y^(2)+6x^(2)y^(3) f(x, y)=5 x^4 y^2+6 x^2 y^3 f ( x , y ) = 5 x 4 y 2 + 6 x 2 y 3 at the point
(
1
,
−
1
)
(
1
,
−
1
)
(1,-1) (1,-1) ( 1 , − 1 ) .
c) State Euler’s theorem for homogeneous functions and verify it for the function:
u
=
x
3
+
y
3
x
+
y
u
=
x
3
+
y
3
x
+
y
u=(x^(3)+y^(3))/(x+y) u=\frac{x^3+y^3}{x+y} u = x 3 + y 3 x + y
a) The population of a town grows at a rate proportional to the population at any time. Its initial population of 500 increases by
15
%
15
%
15% 15 \% 15 %
(
0
,
0
)
(
0
,
0
)
(0,0) (0,0) ( 0 , 0 )
f
(
x
,
y
)
=
{
y
sin
1
x
+
x
sin
1
y
,
x
≠
0
,
y
≠
0
1
,
otherwise
f
(
x
,
y
)
=
y
sin
1
x
+
x
sin
1
y
,
x
≠
0
,
y
≠
0
1
,
otherwise
f(x,y)={[y sin((1)/(x))+x sin((1)/(y))”,”,x!=0″,”y!=0],[1″,”,” otherwise “]:} f(x, y)=\left\{\begin{array}{cc}
y \sin \frac{1}{x}+x \sin \frac{1}{y}, & x \neq 0, y \neq 0 \\
1, & \text { otherwise }
\end{array}\right. f ( x , y ) = { y sin 1 x + x sin 1 y , x ≠ 0 , y ≠ 0 1 , otherwise Answer:
Question:-1(a)
Solve the differential equation:
d
y
d
x
=
−
2
x
+
y
+
1
4
x
+
2
y
−
1
d
y
d
x
=
−
2
x
+
y
+
1
4
x
+
2
y
−
1
(dy)/(dx)=-(2x+y+1)/(4x+2y-1) \frac{d y}{d x} = -\frac{2 x + y + 1}{4 x + 2 y – 1} d y d x = − 2 x + y + 1 4 x + 2 y − 1 Answer:
To solve the differential equation
d
y
d
x
=
−
2
x
+
y
+
1
4
x
+
2
y
−
1
d
y
d
x
=
−
2
x
+
y
+
1
4
x
+
2
y
−
1
(dy)/(dx)=-(2x+y+1)/(4x+2y-1) \frac{dy}{dx} = -\frac{2x + y + 1}{4x + 2y – 1} d y d x = − 2 x + y + 1 4 x + 2 y − 1 , we observe that it is of the form
d
y
d
x
=
f
(
x
,
y
)
d
y
d
x
=
f
(
x
,
y
)
(dy)/(dx)=f(x,y) \frac{dy}{dx} = f(x, y) d y d x = f ( x , y ) . The right-hand side suggests a possible homogeneous structure, so let’s explore that.
First, rewrite the equation. The numerator is
2
x
+
y
+
1
2
x
+
y
+
1
2x+y+1 2x + y + 1 2 x + y + 1 , and the denominator is
4
x
+
2
y
−
1
4
x
+
2
y
−
1
4x+2y-1 4x + 2y – 1 4 x + 2 y − 1 . Notice that the coefficients suggest a linear form. Let’s test if the equation is homogeneous by examining the function:
f
(
x
,
y
)
=
−
2
x
+
y
+
1
4
x
+
2
y
−
1
f
(
x
,
y
)
=
−
2
x
+
y
+
1
4
x
+
2
y
−
1
f(x,y)=-(2x+y+1)/(4x+2y-1) f(x, y) = -\frac{2x + y + 1}{4x + 2y – 1} f ( x , y ) = − 2 x + y + 1 4 x + 2 y − 1 To check for homogeneity, substitute
x
→
t
x
x
→
t
x
x rarr tx x \to tx x → t x ,
y
→
t
y
y
→
t
y
y rarr ty y \to ty y → t y :
Numerator:
2
(
t
x
)
+
(
t
y
)
+
1
=
2
t
x
+
t
y
+
1
2
(
t
x
)
+
(
t
y
)
+
1
=
2
t
x
+
t
y
+
1
2(tx)+(ty)+1=2tx+ty+1 2(tx) + (ty) + 1 = 2tx + ty + 1 2 ( t x ) + ( t y ) + 1 = 2 t x + t y + 1 Denominator:
4
(
t
x
)
+
2
(
t
y
)
−
1
=
4
t
x
+
2
t
y
−
1
4
(
t
x
)
+
2
(
t
y
)
−
1
=
4
t
x
+
2
t
y
−
1
4(tx)+2(ty)-1=4tx+2ty-1 4(tx) + 2(ty) – 1 = 4tx + 2ty – 1 4 ( t x ) + 2 ( t y ) − 1 = 4 t x + 2 t y − 1 So,
f
(
t
x
,
t
y
)
=
−
2
t
x
+
t
y
+
1
4
t
x
+
2
t
y
−
1
f
(
t
x
,
t
y
)
=
−
2
t
x
+
t
y
+
1
4
t
x
+
2
t
y
−
1
f(tx,ty)=-(2tx+ty+1)/(4tx+2ty-1) f(tx, ty) = -\frac{2tx + ty + 1}{4tx + 2ty – 1} f ( t x , t y ) = − 2 t x + t y + 1 4 t x + 2 t y − 1 This is not equal to
f
(
x
,
y
)
f
(
x
,
y
)
f(x,y) f(x, y) f ( x , y ) due to the constant terms
+
1
+
1
+1 +1 + 1 and
−
1
−
1
-1 -1 − 1 , indicating the equation is not homogeneous in the standard sense. However, the structure resembles a linear or exact form, so let’s try a substitution to simplify it.
Notice the numerator and denominator are linear in
x
x
x x x and
y
y
y y y . Let’s try a substitution that aligns the variables. Define:
u
=
2
x
+
y
u
=
2
x
+
y
u=2x+y u = 2x + y u = 2 x + y Then,
y
=
u
−
2
x
y
=
u
−
2
x
y=u-2x y = u – 2x y = u − 2 x , and differentiate with respect to
x
x
x x x :
d
y
d
x
=
d
u
d
x
−
2
d
y
d
x
=
d
u
d
x
−
2
(dy)/(dx)=(du)/(dx)-2 \frac{dy}{dx} = \frac{du}{dx} – 2 d y d x = d u d x − 2 Now, compute the numerator and denominator in terms of
u
u
u u u :
Numerator:
2
x
+
y
+
1
=
2
x
+
(
u
−
2
x
)
+
1
=
u
+
1
2
x
+
y
+
1
=
2
x
+
(
u
−
2
x
)
+
1
=
u
+
1
2x+y+1=2x+(u-2x)+1=u+1 2x + y + 1 = 2x + (u – 2x) + 1 = u + 1 2 x + y + 1 = 2 x + ( u − 2 x ) + 1 = u + 1 Denominator:
4
x
+
2
y
−
1
=
4
x
+
2
(
u
−
2
x
)
−
1
=
4
x
+
2
u
−
4
x
−
1
=
2
u
−
1
4
x
+
2
y
−
1
=
4
x
+
2
(
u
−
2
x
)
−
1
=
4
x
+
2
u
−
4
x
−
1
=
2
u
−
1
4x+2y-1=4x+2(u-2x)-1=4x+2u-4x-1=2u-1 4x + 2y – 1 = 4x + 2(u – 2x) – 1 = 4x + 2u – 4x – 1 = 2u – 1 4 x + 2 y − 1 = 4 x + 2 ( u − 2 x ) − 1 = 4 x + 2 u − 4 x − 1 = 2 u − 1 The differential equation becomes:
d
u
d
x
−
2
=
−
u
+
1
2
u
−
1
d
u
d
x
−
2
=
−
u
+
1
2
u
−
1
(du)/(dx)-2=-(u+1)/(2u-1) \frac{du}{dx} – 2 = -\frac{u + 1}{2u – 1} d u d x − 2 = − u + 1 2 u − 1 Rearrange:
d
u
d
x
=
2
−
u
+
1
2
u
−
1
d
u
d
x
=
2
−
u
+
1
2
u
−
1
(du)/(dx)=2-(u+1)/(2u-1) \frac{du}{dx} = 2 – \frac{u + 1}{2u – 1} d u d x = 2 − u + 1 2 u − 1 Combine terms over a common denominator:
2
=
2
(
2
u
−
1
)
2
u
−
1
2
=
2
(
2
u
−
1
)
2
u
−
1
2=(2(2u-1))/(2u-1) 2 = \frac{2(2u – 1)}{2u – 1} 2 = 2 ( 2 u − 1 ) 2 u − 1 So,
2
−
u
+
1
2
u
−
1
=
2
(
2
u
−
1
)
−
(
u
+
1
)
2
u
−
1
=
4
u
−
2
−
u
−
1
2
u
−
1
=
3
u
−
3
2
u
−
1
=
3
(
u
−
1
)
2
u
−
1
2
−
u
+
1
2
u
−
1
=
2
(
2
u
−
1
)
−
(
u
+
1
)
2
u
−
1
=
4
u
−
2
−
u
−
1
2
u
−
1
=
3
u
−
3
2
u
−
1
=
3
(
u
−
1
)
2
u
−
1
2-(u+1)/(2u-1)=(2(2u-1)-(u+1))/(2u-1)=(4u-2-u-1)/(2u-1)=(3u-3)/(2u-1)=(3(u-1))/(2u-1) 2 – \frac{u + 1}{2u – 1} = \frac{2(2u – 1) – (u + 1)}{2u – 1} = \frac{4u – 2 – u – 1}{2u – 1} = \frac{3u – 3}{2u – 1} = \frac{3(u – 1)}{2u – 1} 2 − u + 1 2 u − 1 = 2 ( 2 u − 1 ) − ( u + 1 ) 2 u − 1 = 4 u − 2 − u − 1 2 u − 1 = 3 u − 3 2 u − 1 = 3 ( u − 1 ) 2 u − 1 Thus, the equation is:
d
u
d
x
=
3
(
u
−
1
)
2
u
−
1
d
u
d
x
=
3
(
u
−
1
)
2
u
−
1
(du)/(dx)=(3(u-1))/(2u-1) \frac{du}{dx} = \frac{3(u – 1)}{2u – 1} d u d x = 3 ( u − 1 ) 2 u − 1 This is a separable equation. Rewrite it:
d
u
u
−
1
=
3
2
u
−
1
d
x
d
u
u
−
1
=
3
2
u
−
1
d
x
(du)/(u-1)=(3)/(2u-1)dx \frac{du}{u – 1} = \frac{3}{2u – 1} \, dx d u u − 1 = 3 2 u − 1 d x Integrate both sides:
Left-hand side:
∫
d
u
u
−
1
=
ln
|
u
−
1
|
∫
d
u
u
−
1
=
ln
|
u
−
1
|
int(du)/(u-1)=ln |u-1| \int \frac{du}{u – 1} = \ln|u – 1| ∫ d u u − 1 = ln | u − 1 | Right-hand side:
∫
3
2
u
−
1
d
x
∫
3
2
u
−
1
d
x
int(3)/(2u-1)dx \int \frac{3}{2u – 1} \, dx ∫ 3 2 u − 1 d x . Let
v
=
2
u
−
1
v
=
2
u
−
1
v=2u-1 v = 2u – 1 v = 2 u − 1 , so
d
v
=
2
d
u
d
v
=
2
d
u
dv=2du dv = 2 \, du d v = 2 d u ,
d
u
=
d
v
2
d
u
=
d
v
2
du=(dv)/(2) du = \frac{dv}{2} d u = d v 2 . Then:
∫
3
2
u
−
1
d
x
=
∫
3
v
⋅
1
2
d
v
=
3
2
∫
1
v
d
v
=
3
2
ln
|
v
|
=
3
2
ln
|
2
u
−
1
|
∫
3
2
u
−
1
d
x
=
∫
3
v
⋅
1
2
d
v
=
3
2
∫
1
v
d
v
=
3
2
ln
|
v
|
=
3
2
ln
|
2
u
−
1
|
int(3)/(2u-1)dx=int(3)/(v)*(1)/(2)dv=(3)/(2)int(1)/(v)dv=(3)/(2)ln |v|=(3)/(2)ln |2u-1| \int \frac{3}{2u – 1} \, dx = \int \frac{3}{v} \cdot \frac{1}{2} \, dv = \frac{3}{2} \int \frac{1}{v} \, dv = \frac{3}{2} \ln|v| = \frac{3}{2} \ln|2u – 1| ∫ 3 2 u − 1 d x = ∫ 3 v ⋅ 1 2 d v = 3 2 ∫ 1 v d v = 3 2 ln | v | = 3 2 ln | 2 u − 1 | Since
u
u
u u u is a function of
x
x
x x x , the right-hand side is
3
2
ln
|
2
u
−
1
|
=
∫
3
2
u
−
1
d
u
3
2
ln
|
2
u
−
1
|
=
∫
3
2
u
−
1
d
u
(3)/(2)ln |2u-1|=int(3)/(2u-1)du \frac{3}{2} \ln|2u – 1| = \int \frac{3}{2u – 1} \, du 3 2 ln | 2 u − 1 | = ∫ 3 2 u − 1 d u , and we treat
x
x
x x x implicitly. The equation becomes:
ln
|
u
−
1
|
=
3
2
ln
|
2
u
−
1
|
+
C
ln
|
u
−
1
|
=
3
2
ln
|
2
u
−
1
|
+
C
ln |u-1|=(3)/(2)ln |2u-1|+C \ln|u – 1| = \frac{3}{2} \ln|2u – 1| + C ln | u − 1 | = 3 2 ln | 2 u − 1 | + C Exponentiate both sides:
|
u
−
1
|
=
e
C
|
2
u
−
1
|
3
/
2
|
u
−
1
|
=
e
C
|
2
u
−
1
|
3
/
2
|u-1|=e^(C)|2u-1|^(3//2) |u – 1| = e^C |2u – 1|^{3/2} | u − 1 | = e C | 2 u − 1 | 3 / 2 Let
k
=
e
C
k
=
e
C
k=e^(C) k = e^C k = e C , a positive constant, and handle absolute values by considering signs later:
u
−
1
=
±
k
(
2
u
−
1
)
3
/
2
u
−
1
=
±
k
(
2
u
−
1
)
3
/
2
u-1=+-k(2u-1)^(3//2) u – 1 = \pm k (2u – 1)^{3/2} u − 1 = ± k ( 2 u − 1 ) 3 / 2 Substitute back
u
=
2
x
+
y
u
=
2
x
+
y
u=2x+y u = 2x + y u = 2 x + y :
2
x
+
y
−
1
=
±
k
(
4
x
+
2
y
−
1
)
3
/
2
2
x
+
y
−
1
=
±
k
(
4
x
+
2
y
−
1
)
3
/
2
2x+y-1=+-k(4x+2y-1)^(3//2) 2x + y – 1 = \pm k (4x + 2y – 1)^{3/2} 2 x + y − 1 = ± k ( 4 x + 2 y − 1 ) 3 / 2 The solution is:
2
x
+
y
−
1
(
4
x
+
2
y
−
1
)
3
/
2
=
C
2
x
+
y
−
1
(
4
x
+
2
y
−
1
)
3
/
2
=
C
(2x+y-1)/((4x+2y-1)^(3//2))=C \boxed{\frac{2x + y – 1}{(4x + 2y – 1)^{3/2}} = C} 2 x + y − 1 ( 4 x + 2 y − 1 ) 3 / 2 = C where
C
C
C C C is an arbitrary constant.
Question:-1(b)
Solve the simultaneous equations:
d
x
y
−
z
=
d
y
z
−
x
=
d
z
x
−
y
d
x
y
−
z
=
d
y
z
−
x
=
d
z
x
−
y
(dx)/(y-z)=(dy)/(z-x)=(dz)/(x-y) \frac{d x}{y – z} = \frac{d y}{z – x} = \frac{d z}{x – y} d x y − z = d y z − x = d z x − y Answer:
To solve the simultaneous differential equations:
d
x
y
−
z
=
d
y
z
−
x
=
d
z
x
−
y
,
d
x
y
−
z
=
d
y
z
−
x
=
d
z
x
−
y
,
(dx)/(y-z)=(dy)/(z-x)=(dz)/(x-y), \frac{dx}{y – z} = \frac{dy}{z – x} = \frac{dz}{x – y}, d x y − z = d y z − x = d z x − y , we aim to find relationships between
x
x
x x x ,
y
y
y y y , and
z
z
z z z that satisfy all equations.
Solution:
First Integral:
Consider the multipliers
1
,
1
,
1
1
,
1
,
1
1,1,1 1, 1, 1 1 , 1 , 1 for the numerators
d
x
,
d
y
,
d
z
d
x
,
d
y
,
d
z
dx,dy,dz dx, dy, dz d x , d y , d z . Multiply each term of the given equations accordingly:
d
x
y
−
z
=
d
y
z
−
x
=
d
z
x
−
y
=
k
.
d
x
y
−
z
=
d
y
z
−
x
=
d
z
x
−
y
=
k
.
(dx)/(y-z)=(dy)/(z-x)=(dz)/(x-y)=k. \frac{dx}{y – z} = \frac{dy}{z – x} = \frac{dz}{x – y} = k. d x y − z = d y z − x = d z x − y = k . This gives:
d
x
=
k
(
y
−
z
)
,
d
y
=
k
(
z
−
x
)
,
d
z
=
k
(
x
−
y
)
.
d
x
=
k
(
y
−
z
)
,
d
y
=
k
(
z
−
x
)
,
d
z
=
k
(
x
−
y
)
.
dx=k(y-z),quad dy=k(z-x),quad dz=k(x-y). dx = k(y – z), \quad dy = k(z – x), \quad dz = k(x – y). d x = k ( y − z ) , d y = k ( z − x ) , d z = k ( x − y ) . Sum the numerators:
d
x
+
d
y
+
d
z
=
k
(
y
−
z
+
z
−
x
+
x
−
y
)
=
k
(
0
)
=
0.
d
x
+
d
y
+
d
z
=
k
(
y
−
z
+
z
−
x
+
x
−
y
)
=
k
(
0
)
=
0.
dx+dy+dz=k(y-z+z-x+x-y)=k(0)=0. dx + dy + dz = k(y – z + z – x + x – y) = k(0) = 0. d x + d y + d z = k ( y − z + z − x + x − y ) = k ( 0 ) = 0. Thus:
d
x
+
d
y
+
d
z
=
0.
d
x
+
d
y
+
d
z
=
0.
dx+dy+dz=0. dx + dy + dz = 0. d x + d y + d z = 0. Integrate this differential:
x
+
y
+
z
=
c
1
,
x
+
y
+
z
=
c
1
,
x+y+z=c_(1), x + y + z = c_1, x + y + z = c 1 , where
c
1
c
1
c_(1) c_1 c 1 is an arbitrary constant. This is the first integral, indicating that the sum of
x
,
y
,
z
x
,
y
,
z
x,y,z x, y, z x , y , z is constant.
Second Integral:
Now, choose multipliers
x
,
y
,
z
x
,
y
,
z
x,y,z x, y, z x , y , z for
d
x
,
d
y
,
d
z
d
x
,
d
y
,
d
z
dx,dy,dz dx, dy, dz d x , d y , d z . Multiply each term:
x
⋅
d
x
y
−
z
=
y
⋅
d
y
z
−
x
=
z
⋅
d
z
x
−
y
=
m
.
x
⋅
d
x
y
−
z
=
y
⋅
d
y
z
−
x
=
z
⋅
d
z
x
−
y
=
m
.
x*(dx)/(y-z)=y*(dy)/(z-x)=z*(dz)/(x-y)=m. x \cdot \frac{dx}{y – z} = y \cdot \frac{dy}{z – x} = z \cdot \frac{dz}{x – y} = m. x ⋅ d x y − z = y ⋅ d y z − x = z ⋅ d z x − y = m . This yields:
x
d
x
=
m
(
y
−
z
)
,
y
d
y
=
m
(
z
−
x
)
,
z
d
z
=
m
(
x
−
y
)
.
x
d
x
=
m
(
y
−
z
)
,
y
d
y
=
m
(
z
−
x
)
,
z
d
z
=
m
(
x
−
y
)
.
xdx=m(y-z),quad ydy=m(z-x),quad zdz=m(x-y). x dx = m(y – z), \quad y dy = m(z – x), \quad z dz = m(x – y). x d x = m ( y − z ) , y d y = m ( z − x ) , z d z = m ( x − y ) . Sum the numerators:
x
d
x
+
y
d
y
+
z
d
z
.
x
d
x
+
y
d
y
+
z
d
z
.
xdx+ydy+zdz. x dx + y dy + z dz. x d x + y d y + z d z . Compute the denominator after scaling:
x
(
y
−
z
)
+
y
(
z
−
x
)
+
z
(
x
−
y
)
=
x
y
−
x
z
+
y
z
−
y
x
+
z
x
−
z
y
=
0.
x
(
y
−
z
)
+
y
(
z
−
x
)
+
z
(
x
−
y
)
=
x
y
−
x
z
+
y
z
−
y
x
+
z
x
−
z
y
=
0.
x(y-z)+y(z-x)+z(x-y)=xy-xz+yz-yx+zx-zy=0. x(y – z) + y(z – x) + z(x – y) = xy – xz + yz – yx + zx – zy = 0. x ( y − z ) + y ( z − x ) + z ( x − y ) = x y − x z + y z − y x + z x − z y = 0. Thus:
x
d
x
+
y
d
y
+
z
d
z
=
m
⋅
0
=
0.
x
d
x
+
y
d
y
+
z
d
z
=
m
⋅
0
=
0.
xdx+ydy+zdz=m*0=0. x dx + y dy + z dz = m \cdot 0 = 0. x d x + y d y + z d z = m ⋅ 0 = 0. So:
x
d
x
+
y
d
y
+
z
d
z
=
0.
x
d
x
+
y
d
y
+
z
d
z
=
0.
xdx+ydy+zdz=0. x dx + y dy + z dz = 0. x d x + y d y + z d z = 0. Rewrite as:
1
2
(
2
x
d
x
+
2
y
d
y
+
2
z
d
z
)
=
0.
1
2
(
2
x
d
x
+
2
y
d
y
+
2
z
d
z
)
=
0.
(1)/(2)(2xdx+2ydy+2zdz)=0. \frac{1}{2} (2x dx + 2y dy + 2z dz) = 0. 1 2 ( 2 x d x + 2 y d y + 2 z d z ) = 0. Integrate:
∫
(
2
x
d
x
+
2
y
d
y
+
2
z
d
z
)
=
2
∫
x
d
x
+
2
∫
y
d
y
+
2
∫
z
d
z
=
2
⋅
x
2
2
+
2
⋅
y
2
2
+
2
⋅
z
2
2
=
x
2
+
y
2
+
z
2
.
∫
(
2
x
d
x
+
2
y
d
y
+
2
z
d
z
)
=
2
∫
x
d
x
+
2
∫
y
d
y
+
2
∫
z
d
z
=
2
⋅
x
2
2
+
2
⋅
y
2
2
+
2
⋅
z
2
2
=
x
2
+
y
2
+
z
2
.
int(2xdx+2ydy+2zdz)=2int xdx+2int ydy+2int zdz=2*(x^(2))/(2)+2*(y^(2))/(2)+2*(z^(2))/(2)=x^(2)+y^(2)+z^(2). \int (2x dx + 2y dy + 2z dz) = 2 \int x dx + 2 \int y dy + 2 \int z dz = 2 \cdot \frac{x^2}{2} + 2 \cdot \frac{y^2}{2} + 2 \cdot \frac{z^2}{2} = x^2 + y^2 + z^2. ∫ ( 2 x d x + 2 y d y + 2 z d z ) = 2 ∫ x d x + 2 ∫ y d y + 2 ∫ z d z = 2 ⋅ x 2 2 + 2 ⋅ y 2 2 + 2 ⋅ z 2 2 = x 2 + y 2 + z 2 . Thus:
x
2
+
y
2
+
z
2
=
c
2
,
x
2
+
y
2
+
z
2
=
c
2
,
x^(2)+y^(2)+z^(2)=c_(2), x^2 + y^2 + z^2 = c_2, x 2 + y 2 + z 2 = c 2 , where
c
2
c
2
c_(2) c_2 c 2 is another arbitrary constant. This is the second integral, indicating that the sum of the squares of
x
,
y
,
z
x
,
y
,
z
x,y,z x, y, z x , y , z is constant.
Final Solution:
x
+
y
+
z
=
c
1
,
x
2
+
y
2
+
z
2
=
c
2
,
x
+
y
+
z
=
c
1
,
x
2
+
y
2
+
z
2
=
c
2
,
x+y+z=c_(1),quadx^(2)+y^(2)+z^(2)=c_(2), x + y + z = c_1, \quad x^2 + y^2 + z^2 = c_2, x + y + z = c 1 , x 2 + y 2 + z 2 = c 2 , where
c
1
c
1
c_(1) c_1 c 1 and
c
2
c
2
c_(2) c_2 c 2 are arbitrary constants determined by initial conditions. These equations describe the relationships between
x
,
y
,
z
x
,
y
,
z
x,y,z x, y, z x , y , z that satisfy the original system.