(dy/dx)^3-2xy(dy/dx)+4y^2=0
Answers
Answer:
Step 1: Change the given equation into homogenous differential equation.
dx
dy
=
(x+y)
2
x
2
−2xy−y
2
Divide numerator and denominator x
2
, we get
dx
dy
=
(1+
x
y
)
2
1−
x
2y
−
x
2
y
2
Step 2: Substitute y = vx to change the above differential.
y=vx ⇒
dx
dy
=v+x
dx
dv
v+x
dx
dv
=
(1+v)
2
2−(v+1)
2
x
dx
dv
=
(1+v)
2
2−(v+1)
2
−v(1+v)
2
x
dx
dv
=
(1+v)
2
−v
3
−3v
2
−3v+1
v
3
+3v
2
+3v−1
v
2
+2v+1
dv=
x
−dx
Step 3: Integrate both sides.
∫
v
3
+3v
2
+3v−1
v
2
+2v+1
dv=∫
x
−dx
Put v
3
+3v
2
+3v−1=t⇒(3v
2
+6v+3)dv=dt
∫
3t
dt
=∫
x
−dx
3
1
logt=−logx+c [∵∫
x
dx
=logx+c]
log(v
3
+3v
2
+3v−1)+logx=c
log(
x
3
y
3
+
x
2
3y
2
+
x
3y
−1)x=c
(
x
2
y
3
+
x
3y
2
+3y−1)=c
′