(2x + y - 3) dy = (x + 2y - 3) dx.
Answers
Answer:
Answers
Steve M
Apr 7, 2018
y
+
x
−
2
=
A
(
y
−
x
)
3
Explanation:
We have:
d
y
d
x
=
x
+
2
y
−
3
2
x
+
y
−
3
..... [A]
Our standard toolkit for DE's cannot be used. However we can perform a transformation to remove the constants from the linear numerator and denominator.
Consider the simultaneous equations
{
x
+
2
y
−
3
=
0
2
x
+
y
−
3
=
0
⇒
{
x
=
1
y
=
1
As a result we perform two linear transformations:
Let
{
u
=
x
−
1
v
=
y
−
1
⇔
{
x
=
u
+
1
y
=
v
+
1
⇒
⎧
⎨
⎩
d
x
d
u
=
1
d
y
d
v
=
1
And if we substitute into the DE [A] we get
d
y
d
x
=
(
u
+
1
)
+
2
(
v
+
1
)
−
3
2
(
u
+
1
)
+
(
v
+
1
)
−
3
=
u
+
1
+
2
v
+
2
−
3
2
u
+
2
+
v
+
1
−
3
=
u
+
2
v
2
u
+
v
And utilising the chain rule we have:
d
y
d
x
=
d
y
d
v
d
v
d
u
d
u
d
x
⇒
d
y
d
x
=
d
v
d
u
Thus we have a transformed equation
d
v
d
u
=
u
+
2
v
2
u
+
v
..... [B]
This is now in a form that we can handle using a substitution of the form
v
=
w
u
which if we differentiate wrt
u
using the product gives us:
d
v
d
u
=
(
w
)
(
d
d
u
u
)
+
(
d
d
u
w
)
(
u
)
=
w
+
u
d
w
d
u
Using this substitution into our modified DE [B] we get:
w
+
u
d
w
d
u
=
u
+
2
w
u
2
u
+
w
u
∴
w
+
u
d
w
d
u
=
u
+
2
w
u
2
u
+
w
u
∴
u
d
w
d
u
=
u
+
2
w
u
2
u
+
w
u
−
w
∴
u
d
w
d
u
=
(
u
+
2
w
u
)
−
w
(
2
u
+
w
u
)
2
u
+
w
u
∴
u
d
w
d
u
=
u
+
2
w
u
−
2
u
w
−
w
2
u
2
u
+
w
u
∴
u
d
w
d
u
=
u
(
1
−
w
2
)
u
(
2
+
w
)
∴
u
d
w
d
u
=
1
−
w
2
2
+
w