(x²-3xy^2) dx +(3x^2y-y^3).dy= 0 solve the differential equations
Answers
Answer:
⇒
dx
dy
=−
(y
3
+3x
2
y)
(x
3
+3xy
2
)
=−
3x
2
y(
3x
2
y
y
3
+1)
3xy
2
(
3xy
2
x
3
+1)
=−
x(
3x
2
y
2
+1)
y(
3y
2
x
2
+1)
⇒
dx
dy
=f(
x
y
)
⇒ the given differential equation is a homogeneous equation.
The solution of the given differential equation is :
Put y=vx
⇒
dx
dy
=v+x
dx
dv
v+x
dx
dv
=−
x(
3x
2
(vx)
2
)+1
vx(
3(vx)
2
x
2
+1)
=−v
(
3
(v)
2
+1)
(
3(v)
2
1
+1)
=−
3+(v)
2
1+3(v)
2
×
v
1
=−
3v+(v)
3
1+3(v)
2
⇒x
dx
dv
=−
3v+(v)
3
1+3(v)
2
−v=−
3v+(v)
3
1+3(v)
2
+3(v)
2
+(v)
4
=
3v+(v)
3
1+6(v)
2
+(v)
4
⇒
1+6(v)
2
+(v)
4
3v+(v)
3
dv=−
x
dx
Integrating both the sides we get:
⇒∫
1+6(v)
2
+(v)
4
3v+(v)
3
dv=−∫
x
dx
+c
As we know that,
dv
d
(1+6(v)
2
+(v)
4
)=12v+4v
3
=4(3v+v
3
)
⇒
4
ln∣1+6(v)
2
+(v)
4
+ln∣x∣=ln∣c∣
Resubstituting the value of y=vx we get
⇒
4
ln∣1+6(
x
y
)
2
+(
x
y
)
4
+ln∣x∣=ln∣c∣
⇒y
4
+6x
2
y
2
+x
3
4=C
Ans :⇒y
4
+6x
2
y
2
+x
3
4=C⇒
dx
dy
=−
(y
3
+3x
2
y)
(x
3
+3xy
2
)
=−
3x
2
y(
3x
2
y
y
3
+1)
3xy
2
(
3xy
2
x
3
+1)
=−
x(
3x
2
y
2
+1)
y(
3y
2
x
2
+1)
⇒
dx
dy
=f(
x
y
)
⇒ the given differential equation is a homogeneous equation.
The solution of the given differential equation is :
Put y=vx
⇒
dx
dy
=v+x
dx
dv
v+x
dx
dv
=−
x(
3x
2
(vx)
2
)+1
vx(
3(vx)
2
x
2
+1)
=−v
(
3
(v)
2
+1)
(
3(v)
2
1
+1)
=−
3+(v)
2
1+3(v)
2
×
v
1
=−
3v+(v)
3
1+3(v)
2
⇒x
dx
dv
=−
3v+(v)
3
1+3(v)
2
−v=−
3v+(v)
3
1+3(v)
2
+3(v)
2
+(v)
4
=
3v+(v)
3
1+6(v)
2
+(v)
4
⇒
1+6(v)
2
+(v)
4
3v+(v)
3
dv=−
x
dx
Integrating both the sides we get:
⇒∫
1+6(v)
2
+(v)
4
3v+(v)
3
dv=−∫
x
dx
+c
As we know that,
dv
d
(1+6(v)
2
+(v)
4
)=12v+4v
3
=4(3v+v
3
)
⇒
4
ln∣1+6(v)
2
+(v)
4
+ln∣x∣=ln∣c∣
Resubstituting the value of y=vx we get
⇒
4
ln∣1+6(
x
y
)
2
+(
x
y
)
4
+ln∣x∣=ln∣c∣
⇒y
4
+6x
2
y
2
+x
3
4=C
Ans :⇒y
4
+6x
2
y
2
+x
3
4=C
Step-by-step explanation: