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Quiz 1 Solutions, Version B, MAP 2302, Fall 2016
1. Determine whether the function y = sin(2x) is a solution to the equation
y 0 + xy = cos(2x) .
Solution: The substitution of y = sin(2x) and y 0 = 2 cos(2x) into the equation yields
2 cos(2x) + x sin(2x) = cos(2x)
⇒
x sin(2x) = − cos(2x)
which is impossible for all x from any open interval. Therefore the said function is not a solution to the given differential equation.
2. Determine for which values of m the function y = xm is a solution of the equation
x2 y 00 − xy 0 − 5y = 0 .
Solution: The substitution of y = xm , y 0 = mxm−1 , and y 00 = m(m−1)xm−2 into the equation
yields
m(m − 1)xm − mxm − 5xm = 0 ⇒ m2 − 2m − 5 = 0
m
because the first equality must hold
The latter quadratic
√
√ for all x (if y = x is a solution).
1± 6
equation has two roots m = 1 ± 6. So the functions y = x
are solutions to the given
differential equation.
3. Find all solutions to the separable equation
q
y 0 = 2 y + 1 cos(x) .
Solution: The right side vanishes if y = −1. Therefore y = −1 is a solution. The other
solutions are found by separating variables (if y 6= −1):
q
dy
= 2 y − 1 cos(x)
dx
⇒
dy
√
= cos(x)dx
2 y−1
⇒
q
⇒
y = −1 + sin(x) + C
y − 1 = sin(x) + C
2
for any choice of a real number C. Note that the above solution does not coincide with the
constant solution y = −1 for any choice of the integration constant C.
4. Find the general solution to the linear equation
xy 0 + 2y = 5x3
Solution: It follows from the standard form of the equation y 0 + py = q that p = 2/x.
Therefore the general solution is sought in the form
y = ze−
R
pdx
= ze−2
R
dx/x
= ze−2 ln x =
z
x2
⇒
y0 =
z0
2z
−
x2 x3
The substitution into the equation yields the following equation equation for the unknown
function z:
!
z0
2z
2z
x 2 − 3 + 2 = 5x3 ⇒ z 0 = 5x4 ⇒ z = x5 + C
x
x
x
Thus, the general solution reads y = z/x2 = x3 + C/x2 .
5. Determine whether the following equation is exact and, if so, find the general solution
to it
y
dx + 1 + ln(x) dy = 0
x
Solution: The equation Mdx + Ndy = 0 is exact if My0 = Nx0 . Here
My0 =
∂ y
1
= ,
∂y x
x
Nx0 =
∂ 1
1 + ln x = = My0
∂x
x
So the equation is exact. Its general solution is a collection of level curves F (x, y) = C where
the function F satisfies the conditions Fx0 = M and Fy0 = N. It follows then
Fx0 =
y
x
F (x, y) = y ln(x) + h(y) ,
⇒
x > 0,
for some h(y). The substitution of this F into the second condition yields an equation for h(y):
y ln(x) + h(y)
0
y
= 1 + ln(x)
⇒
h0 (y) = 1
⇒
h(y) = y .
Thus, the general solution reads y ln(x) + y = C.
6. Use suitable substitutions and transformations to reduce the equation with linear coefficients
(−3x + y − 1)dx + (x + y + 3)dy = 0
to a separable equation. Separate the variables but DO NOT integrate!
Solution: Consider the transformation x = u + h, y = v + k where h and k are constants and
u, v are new variables. In this case, dx = du and dy = dv. The constants h and k are chosen
so that the constant terms in the linear coefficients vanish:
(
−3h + k − 1 = 0
h+k+3 = 0
⇒
h = −1 ,
k = −2
In the new variables the equation can be transformed to a homogeneous equation:
(−3u + v)du + (v + u)dv = 0
⇒
v − 3u
(v/u) − 3
z−3
dv
=
=
=
,
du
v+u
(v/u) + 1
z+1
z=
v
u
because v + u = 0 or v = −u is not a solution (and, hence, v + u 6= 0). The substitution v = zu,
v 0 = z 0 u + z converts the homogeneous equation into a separable one:
u
dz
z−3
+z =
du
z+1
⇒
u
dz
−3 − z 2
=
du
1+z
⇒
Z
(1 + z)dz
=−
3 + z2
Z
du
u
where after the integration the substitutions u = x + 1 and z = v/u = (y + 2)/(x + 1) are to
be made.
7. Formulate the initial value problem for a differential equation that describes the following
mixing problem. Determine the type of the differential equation but do not solve the equation.
A brine solution of salt flows at a constant rate of 6 L/min into a tank that initially held
50 L of brine solution in which was dissolved 0.5 kg of salt. The solution inside the tank is
well stirred and flows out of the tank at the same rate. If the concentration of salt in the brine
entering the tank is 0.05 kg/L, determine the mass of salt in the tank after t minutes.
Solution: Let M(t) be the mass of salt in the tank and V (t) = 50 + (6 − 6)t = 50 L is
the volume of the solution at time t (it does not change with time). Then at a time t, M(t) is
increased by 6 · 0.05 · dt kg during the time interval dt (the salt brought into the tank by the
inflow) and decreased by 6 · n(t) · dt kg, where n(t) = M(t)/V (t) is the output concentration
of salt. Thus, the total mass change is dM = 0.3dt − 6Mdt/50. Therefore the mixing problem
is equivalent to the initial value problem for the linear differential equation:
3
dM
+
M = 0.3 ,
dt
25
M(0) = 0.5 .
8 Extra credit. Use the substitution y = x + 1/v to solve the generalized Riccati equation
y 0 = (y − x)2x3 +
y
x
Solution: The substitution y = x + 1/v, y 0 = 1 − v 0/v 2 into the equation yields
1−
v0
1
1
= 2 x3 + 1 +
2
v
v
xv
⇒
−v 0 = x3 +
v
x
⇒
v0 +
1
v = −x3 ,
x
which is a linear equation that is solved by the substitution
v = ze−
R
dx/x
= ze− ln x =
z0
z
1 z
− 2+
= −x3
x
x
xx
z
x
⇒
⇒
v0 =
z 0 = −x4
z0
z
− 2
x
x
⇒
1
z = − x5 + C
5
⇒
v=
z
1
C
= − x4 +
x
5
x
So the general solution of the Riccati equation is y = x + 1/v where v is given by the above
expression.