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Differential Equations test one solutions
Tuesday, April 25, 2006
x
Z
1. Given the function y = 4+
sin s2 ds
2
2
Z
(a) Calculate y(2) = 4+
(b) Calculate
sin s2 ds = 4+0 = 4
2
2. Is the function y = x+
dy
= sin x2
dx
1
dy 2
a solution of
+y = 3+x2 ? Yes because:
x
dx
dy
d
1
1
+ y2 =
x+
+ x+
dx
dx
x
x
1
1
= 1 − 2 + x2 + 2 + 2
x
x
2
= 3 + x2
3. Determine whether each of the following equations is linear or non-linear.
(a)
dy
= t − y2
dt
non-linear
(b) t
dy
= et y−cos t2
dt
(c) y
linear
dy
= 4t − y
dt
non-linear
4. Which of the following equations is exact?
(a)
3x2 − 4y dx+(2y − 4x) dy = 0
(b)
dy 2
+x y = x or (x2 y−x)dx+(1)dy = 0
dx
2
3x − 4y
?
y
= (2y − 4x)x
x2 y − x
?
y
= (1)x
x2 6= 0 not exact
−4 = −4 exact
page two
5. Given the initial value problem
t
dy
3
4
+
y=
and y(3) = 5
dt t − 2
t+1
determine the largest t-interval over which a unique solution is guaranteed to exist. Rewrite
the differential equation as
dy
3
4
+
y=
dt t(t − 2)
t(t + 1)
The coefficients are not continuous at t = 0 and t = 2. The solution to the initial value
problem exists until a discontinuity is reached, so that the solution will exist on the interval
t > 2.
6. Given the initial value problem
dy
√
= y 3 + x 3 y and y(3) = 0
dx
can one be certain that a unique solution exists?
∂ 3
y + xy 1/3 =
y=0
∂y
x
3y + √
3 3 y2
2
!
= undefined
y=0
Therefore a unique solution is not certain to exist.
7. Use Euler’s method (i.e. the tangent-line method) with step-size h = 1/2 to fill in the
following table for the initial value problem
dy
= 2x − y 2 and y(1) = 2
dx
x
y
dy
dx
1
2
2(1) − 22 = −2
(−2)(1.5 − 1) = −1
1.5
2 + (−1) = 1
2(1.5) − (1)2 = 2
(2)(0.5) = 1
∆y ≈
dy
∆x
dx
page three
8. Solve each of the following differential equations or initial value problems:
(a)
dy
= 2xey
dx
separable
=⇒
Z
Z
Z
dy Z
−y
=
2x
dx
=⇒
e
dy
=
2x dx =⇒ −e−y = x2 +C
ey
y = − ln(−x2 + C 0 ) where C 0 = −C
(b)



dy
dt


2t dy

+ 2y = 5e3t 

y(0) = 3
Z
linear
=⇒ find integrating factor: µ = exp
2 dt = e2t . Multiply equation:


Z
d 2t 5t
2t
e
+ 2e y = 5e e =⇒
e y = 5e =⇒ e y = 5e5t dt = e5t + C
dt
dt
2t
2t 3t
Plug in the initial conditions to find that C = 2, and solve for y:
y = e3t + 2e−2t
(c)
3x2 − 4y dx+(2y − 4x) dy = 0
This equation is exact (see problem 4a). A solution is of the form F (x, y) = C, where
F (x, y) =
Z
2
3
3x − 4y dx = x − 4xy + g(y) and F (x, y) =
Therefore the solution is x3 − 4xy + y 2 = C.
Z
2y − 4x dy = y 2 − 4xy + h(x)
page four
9. The rate of population growth of a small country is proportional to the country’s population. At time t = 0 years the country’s population is equal to 5 million.
(a) From time t = 0 years to time t = 20 years the population grows at an instantaneous rate
equal to 4% of the population. Set up the differential equation for the country’s population
P (t) and solve it.
dP
= .04P =⇒ P = 5e.04t when 0 ≤ t ≤ 20
dt
(b) Starting at time t = 20 years the population grows at an instantaneous rate equal to
only 2% of the population. Set up the differential equation for the country’s population
during this interval and solve it.
dP
= .02P =⇒ P = Ce.02t = 5e.04(20) e.02(t−20) = 5e.02(t+20) when t ≥ 20
dt