8. The solution of the differential equation
x^2dy + y(x + y)dx = 0) is
1. x^2y=c^2(2x+y)
2. xy^2=c^2(2x+y)
3. x^2y=c^2(x +2y)
4. xy^2=c^2(x+2y)
Answers
Answered by
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Step-by-step explanation:
x+y−1)dx=(2x+2y−3)dy
−
(2x+2y−3)
(x+y−1)
=
dx
dy
Let x+y=t
1+
dx
dy
=
dx
dt
(On differentiating wrt n)
⇒−
(2t−3)
(t−1)
=
dx
dt
−1
⇒1−
2t−3
t−1
=
dx
dt
⇒
dx
dt
=
2t−3
2t−3−t+1
=
2t−3
t−2
=
2(t−
2
3
)
t−2
⇒2
dx
dt
=
t−
2
3
t−2
∫
(t−2)
(2t−3)
dt=∫dx
∫(
t−2
2t−4+1
)dt=∫dx
⇒∫(
t−2
2t−4
+
t−2
1
)dt=∫dx
⇒∫2dt+∫
t−2
dt
=∫dx
⇒2t+ln∣t−2∣=x+c
⇒2(x+y)+ln∣x+y−2∣=x+c
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