Download Problem1 Solve the given differential equation 2xy dx + (x + 1)dy = 0

MA262 Linear Algebra and Differential Equations
Fall 2013, Purdue
Quiz2
Problem1
Solve the given differential equation
2xy dx + (x2 + 1)dy = 0
SOLUTION
∂M
= 2x
∂y
∂N
= 2x
N = x2 + 1
∂x
My = Nx ⇒ This is an exact diff. equ.
Z
Z
Φ(x, y) =
M dx = 2xydx = x2 y + h(y)
M = 2xy
N =
∂Φ
⇒ x2 + 1 = x2 + h0 (y) ⇒ h0 (y) = 1 ⇒ h(y) = y + C
∂y
Therefore,
Φ(x, y) = x2 y + y + C
(1)
and the solution is
x2 y + y = C
Remark1: this is also a separable equation. Try it with the separable
method!
Remark2: To those who stop at equation (1), please be attentive in
class, not be absent-minded!
Copy Right by Binghe Chen
1
MA262 Linear Algebra and Differential Equations
Fall 2013, Purdue
Problem2
Solve the given initial-value problem
(x − 1)(x − 2)y 00 = y 0 − 1; y(0) = 0, y 0 (0) = −1
SOLUTION
v=
∂y
∂x
∂v
= y 00
∂x
Plug in
(x − 1)(x − 2)v 0 = v − 1
(2)
It is separable.
dv
1
=
dx
v−1
(x − 1)(x − 2)
−1
1
=
+
dx (a lot of mistakes here)
x−1 x−2
ln |v − 1| = − ln |x − 1| + ln |x − 2| + C
x − 2
+C
= ln x − 1
x−2
v−1 = C
x−1
Use the initial condition y 0 (0) = −1
−1 − 1 = C
Therefore,
v=C
Copy Right by Binghe Chen
0−2
= C · 2 ⇒ C = −1
0−1
x−2
2−x
1
+1=
+1=
x−1
x−1
x−1
2
MA262 Linear Algebra and Differential Equations
Fall 2013, Purdue
Recall v = dy/dx, solve the equation:
dy
1
=
dx
x−1
y = ln |x − 1| + C
Use the initial condition y(0) = 0:
0 = ln |0 − 1| + C ⇒ C = 0
However, y = ln |x − 1| can not be a solution, since it is not continuous
on the whole real line. But it is continuous on either (−∞, 1) or (1, ∞).
The initial conditions tell us the interval should contain the point x =
0, so we choose this one as the solution:
y = ln(1 − x)
Remark: Equation (2) is also linear. Try the method for a linear
equation to solve it!
Copy Right by Binghe Chen
3