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Worksheet 1.1
Exercise 1
State the order of the given ordinary differential equation. Determine whether the equation is linear or nonlinear.
1. (1 − x)y ′′ − 4xy ′ + 5y = cos(x)
3
d y
dy 4
2. x dx
3 − ( dx ) + y = 0
3. t5 y 4 − t3 y ′′ + 6y = 0
q
d2 y
dy 2
4. dx2 = 1 + ( dx
)
Exercise 2
Determine whether the given first-order differential equation is linear in the indicated
dependent variable.
1. (y 2 − 1)dx + xdy = 0, in y, in x
2. udv + (v + uv − ueu )du = 0, in v, in u
Exercise 3
Verify that the indicated function is an explicit solution of the given differential equation.
Assume an appropriate interval I of definition for each solution.
−x
1. 2y ′ + y = 0, y = e 2
2. dy
+ 20y = 24, y = 65 − 56 e−20t
dt
3. y ′′ − 2y ′ + y = 0, y = xex
Exercise 4
Verify that the indicated function y = ϕ(x) is an explicit solution of the given first-order
DE. By considering ϕ simply as a function, give its domain. Then by considering ϕ as a
solution of DE, give at least one interval I of definition
√
1. (y − x)y ′ = y − x + 8, y = x + 4 x + 2
−1
2. 2y ′ = y 3 cos(x), y = (1 − sin(x)) 2
2
3. xy ′ + y = x2 , y = x3 + x1
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Exercise 5
Verify that the indicated expression is an implicit solution of the given first-order DE.
Find at least one explicit solution y = ϕ(x) in each case.
1. dX
= (X − 1)(1 − 2X), ln( 2X−1
)=t
dt
X−1
dy
2. dx
= −x
, x2 + y 2 = 5
y
Exercise 6
Find values of m so that the function y = emx is a solution of the given DE.
1. y ′ + 2y = 0
2. y ′′ − 5y ′ + 6y = 0
3. 5y ′ = 2y
Solutions 1.1
Exercise 1
State the order and determine whether each ODE is linear or nonlinear.
1. (1 − x)y ′′ − 4xy ′ + 5y = cos(x)
The highest derivative is y ′′ , so the equation is of order 2. All terms involving y
and its derivatives have the power equal to 1, and coefficients depend only on x.
⇒ Second-order linear ODE.
2. x
d3 y
−
dx3
dy
dx
4
+y =0
The highest derivative is y (3) , so the equation is of order 3. The term (y ′ )4 makes
the equation nonlinear.
⇒ Third-order notlinear ODE.
3. t5 y 4 − t3 y ′′ + 6y = 0
The highest degree of y (4) is 4, the order of the DE is 2.
⇒ Fourth-order not linear ODE.
d2 y
4.
=
dx2
s
1+
dy
dx
2
The highest derivative is y ′′ , so the equation is of order 2. The square root involving
(y ′ )2 makes it nonlinear.
⇒ Second-order not linear ODE.
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Exercise 2
Determine linearity in the indicated dependent variable.
1. (y 2 − 1)dx + xdy = 0
dy
= 0 In x: rewriting as
In y: the term y 2 appears ⇒ notlinear. DE : (y 2 − 1) + x dx
2 dx
(1 − y ) dy + x = 0, the equation is linear in x.
2. u dv + (v + uv − ueu ) du = 0
dv
dv
In v: the equation is linear in v. DE : u du
+(v +uv −ueu ) = u du
+(1+u)v −ueu = 0
( linearity properties are satisfied)
In u: the equation is not linear in u. DE: u + (v + uv − ueu ) du
=0
dv
Exercise 3
Verify that the function is a solution and give an interval of definition.
1. 2y ′ + y = 0, y = e−x/2
y ′ = − 12 e−x/2
Substitution gives 2y ′ + y = 0. Interval of definition: I = R.
2.
dy
+ 20y = 24, y = 65 − 65 e−20t
dt
y ′ = 24e−20t
Substitution verifies the equation. Interval of definition: I = R.
3. y ′′ − 2y ′ + y = 0, y = xex
y ′ = ex + xex ,
y ′′ = 2ex + xex
Substitution gives 0. Interval of definition: I = R.
Exercise 4
Verify solution, give domain, and one solution interval.
√
1. (y − x)y ′ = y − x + 8, y = x + 4 x + 2
2
Domain as a function: x ≥ −2, [−2, + inf) y ′ = 1 + √x+2
Differentiability requires
x > −2. y is a solution must be defined on I and the derivatives are continuous
on I, by substituting the derivatives into the DE, it reduces the DE to an identity.
Interval of solution:
I = (−2, +∞).
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2. 2y ′ = y 3 cos x, y = (1 − sin x)−1/2
Domain as a function:
nπ
o
R\
+ 2kπ = Uk∈Z (π/2 + 2kπ, π/2 + 2(k + 1)π)
2
−3
y ′ = cos(x)
(1−sin(x)) 2 substitute the y’ into the DE verify that the y is a solution,
2
y must reduces the DE into an identity.(Quick check)
One interval of definition: for example, k = 0
π 5π
I=
,
.
2 2
2
3. xy ′ + y = x2 , y = x3 + x1
Domain as a function: R \ {0}. y ′ = 2x
− x12 ( substitute it in the DE to check the
3
y is a soltuion)
One interval of definition:
I = (0, +∞) or (−∞, 0).
Exercise 5
Verify that the indicated expression is an implicit solution of the given first-order DE,
then find at least one explicit solution.
1. Differential equation:
dX
= (X − 1)(1 − 2X)
dt
Implicit solution:
2X − 1
ln
=t
X −1
Verification.
Differentiate implicitly with respect to t:
d
[ln(2X − 1) − ln(X − 1)] = 1
dt
d
dX
dt
[ln(2X − 1) − ln(X − 1)]
=
=1
dX
dt
dt
2
1
dX
−
=1
2X − 1 X − 1 dt
−1
dX
=1
(2X − 1)(X − 1) dt
dX
= −(2X − 1)(X − 1) = (X − 1)(1 − 2X)
dt
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Thus, the implicit expression satisfies the differential equation.
Explicit solution.
Exponentiating both sides and solve for the dependent variable in this example X:
2X − 1
= et
X −1
2X − 1 = et (X − 1)
X(2 − et ) = 1 − et
X(t) =
1 + et
2 − et
—
2. Differential equation:
dy
x
=−
dx
y
Implicit solution:
x2 + y 2 = 5
Verification.
Differentiate implicitly with respect to x:
d
d
d 2
x + y2 =
5
dx
dx
dx
2x + 2y
dy
=0
dx
dy
x
=−
dx
y
Thus, the implicit relation satisfies the differential equation.
Explicit solutions.
Solving for y:
y 2 = 5 − x2
√
y = ± 5 − x2
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Exercise 6
Find values of m such that y = emx is a solution.
1. y ′ + 2y = 0
m + 2 = 0 ⇒ m = −2.
2. y ′′ − 5y ′ + 6y = 0
m2 − 5m + 6 = 0 ⇒ m = 2, 3.
3. 5y ′ = 2y
2
5m = 2 ⇒ m = .
5
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