MATH 219 - Differential Equations Quiz 3 Name Solutions Date ...

MATH 219 - Differential Equations
Quiz 3
Name Solutions Date
State the type of each of the following equations. (Separable, exact or homogeneous) Equations
may be of more than one type; you only need to state one. Then, find the general
solution to each of the equations.
1. 3x 2 y dx = (3x 3 + y 3 )dy Type: homogeneous
Substitution: x = vy; dx = vdy + ydv
3v 2 y 2 y(vdy + ydv) = (3v 3 y 3 + y 3 )dy divide both sides by y 3
3v 3 dy + 3v 2 y dv = (3v 3 + 1)dy
3v 2 y dv = dy
3v 2 dv = 1
y dy
3v 2
1
dv =
y dy
v 3 = ln |y| + C
3 x
= ln y + C
y
2. x 1 − y 2 dx = dy Type: separable
1
x dx =
1 − y2 dy
1
x dx =
1 − y2 dy
1
2 x2 = sin −1 y + C

3. sec 2 x+ 1
y2 +yexy
+ − 2x
y
dx +
sec 2 x + 1
+ yexy
y2 3 + xexy
dy
dx
M(x, y) = sec 2 x + 1
y2 + yexy ; ∂M
∂y
N(x, y) = − 2x
y3 + xexy ; ∂N
∂x
− 2x
+ xexy
y3 = 0 Type: exact
dy = 0
= − 2
y 3 + exy + xye xy
= − 2
y 3 + exy + xye xy
So, the equation is exact.
Therefore there exists a function f(x, y) such that ∂f
∂f
= M(x, y) and = N(x, y)
∂x ∂y
∂f
= M(x, y), so f(x, y) = M(x, y)dx = sec
∂x 2 x + 1
+ yexy dx =
y2 tan x + x
y2 + exy + h(y)
∂f
= N(x, y), so f(x, y) = N(x, y)dy = −
∂y 2x
+ xexy dy =
y3 x
y2 + exy + g(x)
So, h(y) = 0 and g(x) = tan x.
f(x, y) = tan x + x
y 2 + exy .
Solution:
tan x + x
y 2 + exy = C
Page 2