6. What is the value of IS
(1 Point
- 3x²y + 2 x y2 at x = - 1 and y = 1
5
is this answer -5 correct
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Answer:
x
2
y−2xy
2
)dx=(x
3
−3x
2
y)dy
(x
2
y−2xy
2
)dx−(x
3
−3x
2
y)dy=0
Here,
M=x
2
y−2xy
2
⇒
∂y
∂M
=x
2
−4xy
N=−(x
3
−3x
2
y)⇒
∂x
∂N
=6xy−3x
2
∵
∂y
∂M
=
∂x
∂N
Therefore,
Integrating factor =
Mx+Ny
1
=
x(x
2
y−2xy
2
)+y(3x
2
y−x
3
)
1
=
x
2
y
2
1
Multiplying the given equation by the integrating factor, we get
(
x
2
y
2
1
)(x
2
y−2xy
2
)dx−(
x
2
y
2
1
)(x
3
−3x
2
y)dy=0
⇒(
y
1
−
x
2
)dx−(
y
2
x
−
y
3
)dy=0
Now again,
M=(
y
1
−
x
2
)⇒
∂y
∂M
=−
y
2
1
N=−(
y
2
x
−
y
3
)⇒
∂x
∂N
=−
y
2
1
Now the above equation is an exact differential equation.
Therefore,
Solution of the equation is-
∫Mdx+∫(terms in N not containing x)dy=C
⇒∫(
y
1
−
x
2
)dx+∫(
y
3
)dy=C
⇒
y
x
−2logx+3logy=C
⇒
y
x
−logx
2
+logy
3
=C
⇒
y
x
+log(
x
2
y
3
)=C
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