Solution Of First Order Equations Worksheets With Answers - S. Ghorai Page 2
ADVERTISEMENT
1
2
3
4
5S. Ghorai
2
2.2
Homogeneous form
Let the ODE be of the form
y = f (y/x)
In this case, substitution of v = y/x reduces the above ODE to a seprable ODE.
Comment 1: Sometimes, substitution reduces an ODE to the homogeneous form. For
example, if ae = bd, then h and k can be chosen so that x = u + h and y = v + k
reduces the following ODE
ax + by + c
y = F
dx + ey + f
to a homeogeneous ODE. What happens if ae = bd?
Comment 2: Also, an ODE of the form
y = y/x + g(x)h(y/x)
can be reduced to the separable form by substituting v = y/x.
2
2
Example 3. xyy = y
+ 2x
,
y(1) = 2
Solution: Substituting v = y/x we find
2
2
2
v + xv = v + 2/v
y
= 2x
(C + ln x
)
2
2
Using y(1) = 2, we find C = 2. Hence, y = 2x
(1 + ln x
)
3
Exact equation
A first order ODE of the form
M (x, y) dx + N (x, y) dy = 0
(1)
is exact if there exits a function u(x, y) such that
∂u
∂u
M =
and N =
.
∂x
∂y
Then the above ODE can be written as du = 0 and hence the solution becomes u = C.
Theorem 1. Let M and N be defined and continuously differentiable on a rectangle
rectangle R = (x, y) : x
x
< a, y
y
< b . Then (1) is exact if and only if
0
0
∂M/∂y = ∂N/∂x for all (x, y)
R.
Proof: We shall only prove the necessary part. Assume that (1) is exact. Then there
exits a function u(x, y) such that
∂u
∂u
M =
and N =
.
∂x
∂y
Since M and N have continuous first partial derivatives, we have
2
2
∂M
∂
u
∂N
∂
u
=
and
=
.
∂y
∂y∂x
∂x
∂x∂y
Now continuity of 2nd partial derivative implies ∂M/∂y = ∂N/∂x.
ADVERTISEMENT
0 votes
Related Articles
Related forms
Related Categories
Parent category: Education