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dy dx
Re-write the equation as
y−^4 dy dx
and use the substitution u = y−^3 to obtain the linear equation
−
du dx
Consider first the homogeneous linear equation
−
duh dx
which is separable. Solving it gives
uh = Ce^3 x (^2) / 2 ,
where C is an arbitrary constant. Using variation of parameters, the solution of the inhomogeneous equation will be found in the form
u = C(x)e^3 x^2 /^2.
Substitute this formula in the equation for u to get − 1 3
dC dx e^3 x (^2) / 2 = x^3.
Hence
C(x) = − 3
x^3 e−^3 x (^2) / 2 dx =
x^2 d(e−^3 x (^2) / 2 )
= x^2 e−^3 x (^2) / 2 − 2
xe−^3 x (^2) / 2 dx =
x^2 + 2/ 3
e−^3 x (^2) / 2
where K is an arbitrary constant. Thus
u = Ke−^3 x (^2) / 2
Ke−^3 x (^2) / 2
Finally, since y(0) = 1, K = 1/3 and hence
y =
e−^3 x^2 /^2 + 3x^2 + 2 3
dy dx
3 x + y y − x
, y(1) = 2.
Re-write the equation as
(3x + y)dx + (x − y)dy,
which is exact, since ∂(3x + y) ∂y
∂(x − y) ∂x
It follows that the solution y is given implicitly as
F (x, y) = C,
where C is a constant, and F (x, y) is a solution of the system
∂F
∂x = 3x + y
∂F ∂y = x − y
Integrating the second equation with respect to y gives
F (x, y) = xy − y^2 /2 + f (x).
Substitute this formula into the first equation of the system to get
f ′(x) = 3x =⇒ f (x) = 3x^2 /2 =⇒ F (x, y) = 3x^2 /2 + xy − y^2 / 2.
Finally, since y(1) = 2,
C = F (1, 2) = 3/ 2.
Thus 3 x^2 /2 + xy − y^2 /2 = 3/ 2. Solving this last equation for y yields
y = x +
4 x^2 − 3 ,
where the ”+” sign was chosen in view of the initial condition y(1) = 2.
d^2 y dx^2
dy dx
- 5y = 4ex, y(0) = 0, y′(0) = 0.
Denote by Y (s) the Laplace transform of y:
Y (s) = L{y(x)}(s).
Then s^2 Y + 2sY + 5Y =
s − 1
Y =
(s − 1)(s^2 + 2s + 5)
(s − 1)((s + 1)^2 + 2^2 )
Using the method of elementary fractions:
4 (s − 1)((s + 1)^2 + 2^2 )
A
s − 1
B(s + 1) + 2C (s + 1)^2 + 2^2
A(s^2 + 2s + 5) + B(s^2 − 1) + C(2s − 2) = 4; A = −B = −C = 1/ 2. Thus
Y =
s − 1
s + 1 (s + 1)^2 + 2^2
(s + 1)^2 + 2^2
y =
ex^ − e−x^ cos(2x) − e−x^ sin(2x) 2
- (20 pts) Find and classify all equilibria of the autonomous system
dx dt = x + y
dy dt = xy + 1
The equilibrium points are determined by { (^) x + y = 0
xy + 1 = 0 =⇒ x = −y = ± 1.
Thus (1, −1) and (− 1 , 1) are equilibrium points. Consider the lin- earized system at (1, −1):
du dt = u + v
dv dt = −u + v
where
{ (^) u = x − 1
v = y + 1
The poles p 1 , p 2 of the Laplace transforms of u and v satisfy
p 1 + p 2 = 2, p 1 p 2 = 2,
hence p 1 and p 2 are two conjugate complex numbers with positive real parts. Conclude that solutions will spiral out from the equilibrium at (1, −1). The linearized system at (− 1 , 1) is:
du dt = u + v
dv dt = u − v
where
{ (^) u = x + 1
v = y − 1
In this case, the poles p 1 , p 2 of the Laplace transforms of u and v satisfy
p 1 + p 2 = 0, p 1 p 2 = − 2 ,
hence p 1 and p 2 are two real numbers with opposite signs. Conclude that the equilibrium at (− 1 , 1) is a saddle point.
