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Section 5.5 Exercises
Subsection 💡 Conceptual Quiz
Exercises Exercises
📝: Abbreviations.
Differential Equation
Initial Value Problem
1. True or False.
(a) 👍👎.
Every first-order ordinary differential equation can be solved using the method of separation of variables.
True.
False. Only first-order ordinary differential equations that are separable can be solved using the method of separation of variables.
False.
False. Only first-order ordinary differential equations that are separable can be solved using the method of separation of variables.
(b) 👍👎.
If a differential equation is separable, its solution will always be an explicit function of
\(x \text{.}\)
True.
False. The solution to a separable differential equation is sometimes an implicit function of the independent variable.
False.
False. The solution to a separable differential equation is sometimes an implicit function of the independent variable.
(c) 👍👎.
The method of separation of variables can be applied to some second-order differential equations, but not all of them.
True.
False. The method of separation of variables is only applicable to first-order separable differential equations.
False.
False. The method of separation of variables is only applicable to first-order separable differential equations.
(d) 👍👎.
Any differential equation that can be written in the form
\begin{equation*}
h(y) \frac{dy}{dx} = g(x)\text{,}
\end{equation*}
is separable.
True.
True. Any differential equation that can be written in the form
\begin{equation*}
h(y) \frac{dy}{dx} = g(x)
\end{equation*}
can also be rewritten in the separable form
\begin{equation*}
\frac{dy}{dx} = g(x)\cdot \frac{1}{h(y)}\text{.}
\end{equation*}
False.
True. Any differential equation that can be written in the form
\begin{equation*}
h(y) \frac{dy}{dx} = g(x)
\end{equation*}
can also be rewritten in the separable form
\begin{equation*}
\frac{dy}{dx} = g(x)\cdot \frac{1}{h(y)}\text{.}
\end{equation*}
(e) 👍👎.
The differential equation, below, is
separable .
\begin{equation*}
\frac{dy}{dx} = \frac{\cos x}{y}
\end{equation*}
True.
The correct answer is:
A. It is always separable.
This differential equation is the same as
\begin{equation*}
\frac{dy}{dx} = g(x) \cdot (1)\text{,}
\end{equation*}
which shows that the equation is separable since the function of \(y\) can be constant.
The equation is separable because it is first-order and can be written in the separable form
\begin{equation*}
\frac{dy}{dx} = \cos x \left(\frac{1}{y}\right)\text{.}
\end{equation*}
False.
The correct answer is:
A. It is always separable.
This differential equation is the same as
\begin{equation*}
\frac{dy}{dx} = g(x) \cdot (1)\text{,}
\end{equation*}
which shows that the equation is separable since the function of \(y\) can be constant.
The equation is separable because it is first-order and can be written in the separable form
\begin{equation*}
\frac{dy}{dx} = \cos x \left(\frac{1}{y}\right)\text{.}
\end{equation*}
(i) 👍👎.
The differential equation,
\begin{equation*}
x + \frac{dy}{dx} = y\text{,}
\end{equation*}
can be solved using the separation of variables method.
True.
The differential equation is first order, but it is equivalent to
\begin{equation*}
\frac{dy}{dx}=y-x\text{,}
\end{equation*}
which can not be written in the form \(\dfrac{dy}{dx}=f(x)g(y)\text{.}\)
False.
The differential equation is first order, but it is equivalent to
\begin{equation*}
\frac{dy}{dx}=y-x\text{,}
\end{equation*}
which can not be written in the form \(\dfrac{dy}{dx}=f(x)g(y)\text{.}\)
Hint .
Check that the differential equation is first-order and separable.
(j) 👍👎.
The differential equation,
\begin{equation*}
\frac{d^2z}{dt^2} = \cos^2(z)\text{,}
\end{equation*}
can be solved using the separation of variables method.
True.
The differential equation is not first order, so it is not separable and cannot be solved using the separation of variables method.
False.
The differential equation is not first order, so it is not separable and cannot be solved using the separation of variables method.
Hint .
Check that the differential equation is first-order and separable.
(k) 📖❓ SOV and Nonlinear Equations.
The separation of variables method cannot be applied to nonlinear differential equations.
True.
Solving nonlinear differential equations is one of the strengths of the separation of variables method.
False.
Solving nonlinear differential equations is one of the strengths of the separation of variables method.
2. Multiple Choice.
(a) Initial Step.
What is the first step to verify that the differential equation below is separable?
\begin{equation*}
\frac{dy}{dx} + \frac{x}{y^2} = 0
\end{equation*}
Integrating both sides with respect to \(x\text{.}\)
Incorrect. Integration should be performed after verifying that the equation is separable.
Check that the equation is first order.
Correct! The first step in verifying that an equation is separable is to ensure it is first order.
Combine the fractions on the left-side of the equation.
Incorrect. First of all, \(\frac{dy}{dx}\) is not a fraction. Even if it were, this would not be the first step in verifying separability.
Isolate \(\dfrac{dy}{dx}\) on one side of the equation.
Incorrect. Isolating \(\dfrac{dy}{dx}\) is required to verify the equation is separable, but there is one other detail to check first.
(b) Select the Separable Equations.
Click on each of the separable differential equations below.
A differential equation is separable if you can rearrange it to express it as a product of functions involving only \(x\) and \(y\) separately. Look for equations that can be manipulated into this form, even if they do not initially appear separable.
\(\)
\(\quad\ds x^2\frac{dy}{dx} - y = x^2y\) \(\quad\dfrac{dy}{dx} + xy = y^2\)
\(\)
\(\quad\ds \sqrt{x^2+1}\frac{dy}{dx} = y\) \(\quad\ds xy\frac{dy}{dx} = x + y\)
\(\)
\(\quad\ds x\frac{dy}{dx} - e^x = 3\) \(\quad\dfrac{dy}{dx} - x = \dfrac{y^2}{x}\)
\(\)
(c) When is it Separable?
What can you say about any differential equation that can be written in the form
\begin{equation*}
\frac{dy}{dx} = g(x)\text{?}
\end{equation*}
It is always separable.
It is never separable.
It might be separable, depending on \(g(x)\text{.}\)
It is not a valid equation because it lacks a \(y\) term.
(d) Explicit or Implicit?
Which of the following equations represent an
implicit solution to a differential equation?
Assume \(y\) is the dependent variable and \(t\) is the independent variable.
\(\quad y = \tan(t) + c\) No, this is an explicit solution since \(y\) is written as a function of \(t\text{.}\)
\(\quad y^2 + \ln y = t + c\) Correct. You can’t isolate \(y\) without solving a nonlinear equation, so this is an implicit solution.
\(\quad y^2 + 2e^{y} = t^2 - 2\) Yes, this matches the example you just saw and defines \(y\) only implicitly.
\(\quad y = \tan(t + c)\) No, this is an explicit expression for \(y\) as a function of \(t\text{.}\)
(e) When Does SOV Apply?
Which two properties must a differential equation satisfy in order to solve it using the separation of variables method?
The equation must be linear and separable.
Incorrect. Separable is required, but the equation does not need to be linear.
The equation must be first-order and linear.
Incorrect. A first-order equation is required, but the equation does not need to be linear for the separation of variables method to be applied.
The equation must be second-order and separable.
Incorrect. Separability is required, but the equation cannot be second-order for the separation-of-variables method to be applied.
The equation must be first-order and separable.
Correct! The separation of variables method applies only to first-order separable differential equations.
(f) Combining Constants.
Which of the following are valid ways to combine constants when finalizing your solution? Select ALL that apply.
\(\quad \left\{
\begin{align}
|y| \amp = e^{2x + c_1}\\
y \amp = c_2 e^{2x}
\end{align}
\right.\qquad\) where \(c_2=\pm e^{c_1}\text{.}\)
Correct, since
\begin{align*}
|y| \amp = e^{2x + c_1} \\
y \amp = \pm e^{2x + c_1} \\
y \amp = \pm e^{2x} \cdot e^{c_1} \\
y \amp = c_2 e^{2x} \quad \text{where } c_2 = \pm e^{c_1}
\end{align*}
\(\quad \left\{
\begin{align}
\frac{y^2}{2} \amp = \sin(x) + c_1\\
y^2 \amp = 2\sin(x) + c_2
\end{align}
\right.\qquad\) where \(c_2 = 2c_1\text{.}\)
Correct, since
\begin{align*}
\frac{y^2}{2} \amp = \sin(x) + c_1 \\
y^2 \amp = 2\sin(x) + 2c_1 \\
y^2 \amp = 2\sin(x) + c_2 \quad \text{where } c_2 = 2c_1
\end{align*}
\(\quad \left\{
\begin{align}
y^2 \amp = 2\sin(x) + c_2\\
y \amp = \pm\sqrt{2\sin(x)} + c_3
\end{align}
\right.\qquad\) where \(c_3=\pm \sqrt{c_2}\text{.}\)
Incorrect, you should not combine constants here since \(\quad\sqrt{A + B} \neq \sqrt{A} + \sqrt{B}\text{.}\)
Your solution should be written as
\begin{align*}
y^2 \amp = 2\sin(x) + c_2 \\
y \amp = \pm\sqrt{2\sin(x) + c_2}\quad \leftarrow\text{ done}
\end{align*}
\(\quad \left\{
\begin{align}
e^y \amp = 3y + c_1\\
y \amp = \ln(3y) + c_2
\end{align}
\right.\qquad\) where \(c_2= \ln(c_1)\text{.}\)
Incorrect, you should not combine constants here since \(\ln(A + B) \neq \ln(A) + \ln(B)\text{.}\)
Your solution should be
\begin{align*}
e^y \amp = 3y + c_1 \\
y \amp = \ln(3y + c_1)\quad \leftarrow\text{ done}
\end{align*}
\(\quad \left\{
\begin{align}
xy \amp = e^x + c_1\\
y \amp = \frac{e^x}{x} + c_2
\end{align}
\right.\qquad\) where \(c_2=\frac{c_1}{x}\text{.}\)
Incorrect, you should not combine constants here since \(\frac{c_1}{x}\) is not constant. Instead, your solution should be left as
\begin{align*}
xy \amp = e^x + c_1 \\
y \amp = \frac{e^x + c_1}{x} \\
y \amp = \frac{e^x}{x} + \frac{c_1}{x}\quad \leftarrow\text{ done}
\end{align*}
\(\quad \left\{
\begin{align}
\sqrt{y} \amp = c_1e^{3x}\\
y \amp = c_2\, e^{6x}
\end{align}
\right.\qquad\) where \(c_2= c_1^2\text{.}\)
Correct, since
\begin{align*}
\sqrt{y} \amp = c_1e^{3x} \\
y \amp = (c_1e^{3x})^2 \\
y \amp = c_1^2(e^{3x})^2 \\
y \amp = c_2\, e^{6x}\quad \text{where } c_2 = c_1^2
\end{align*}
3. Short-Answer Questions.
(a) Not Separable.
Give an example of a differential equation that is not separable.
(b) Where Does “Separation of Variables” Come From?
Explain why you think “separation of variables” is an appropriate name for the method in this chapter.
(c) What Could go Wrong?
Ignoring the fact that the following equation is not separable, suppose we applied the separation of variables process to
\begin{equation*}
\frac{dy}{dx} = y^2 + x\text{.}
\end{equation*}
(d) Implicit vs. Explicit.
Some differential equations solved using separation of variables yield explicit solutions, while others yield implicit solutions. Give your interpretation of the difference between an implicit and explicit solution. Provide an example of an implicitly and explicitly defined function.
(e) Thinking in Cases.
All first-order linear differential equations with dependent variable, \(y\text{,}\) can be written in the form
\begin{equation*}
\frac{dy}{dx} + P(x) y = Q(x)
\end{equation*}
where \(P(x)\) and \(Q(x)\) are known functions of \(x\text{.}\) What must be true about \(P(x)\) and \(Q(x)\) for a first-order linear differential equation to be separable?
Hint .
Make some general assumptions about
\(P(x)\) and
\(Q(x)\) and think about how you could break this question into different cases about what
\(P(x)\) and
\(Q(x)\) could be to guarantee separability.
Subsection 🏋️♂️ Practice Drills
Exercises Exercises
1. Verify the solution.
Verify that \(z = \tan(t+C)\) is a solution to the initial value problem
\begin{equation*}
z' - 1 = z^2, \quad z(4) = 9
\end{equation*}
then find the particular solution.
2. Show the differential equation is separable.
\begin{equation*}
t - ss' = - s
\end{equation*}
Separable Form. Determine whether the given differential equation is separable or not. As demonstrated in the examples, if the equation is separable, use parentheses to explicitly show the separable form.
3. \(\dfrac{dz}{dt} = \sin(z+t)\) 4. \(s' = t\ln(s^{2t}) + 8t^2\) 5. \(\dfrac{dy}{dx} = 2y^3 + y + 4\) 6. \(\ds y' - xy = 0\) 7. \(\ds y' = x+y\) 8. \(\ds y'' + y' + y = 0\)
Subsection ✍🏻 Problems
Exercises Exercises
Solve the Differential Equations (Step-By-Step). Solve each of the following differential equations using the separation of variables method. Press
Activate to submit your answers and get instant feedback.
1. Determine the Separable Form.
Write each differential equation in the separable form
\begin{equation*}
\frac{dy}{dx} = f(x)\cdot g(y)\text{.}
\end{equation*}
Type your answer for \(f(x)\) into the left box and \(g(y)\) into the right box.
2. Reorder the Steps.
Reorder the steps required to solve the differential equation,
\begin{equation*}
\frac{dy}{dx} - y\cos(x) = y\text{,}
\end{equation*}
using the separation of variables method.
Check for 1st-order.
---
\(\dfrac{dy}{dx} = y + y\cos(x)\)
---
\(\dfrac{dy}{dx} = (1 + \cos(x))y\)
---
\(\dfrac{1}{y} dy = (1 + \cos(x))dx\)
---
\(\ds\int \frac{1}{y} dy =\) \(\qquad\ds\int (1 + \cos(x))dx\)
---
\(\ln|y| = x + \sin(x) + C \)
---
\(y = Ce^{x + \sin(x)} \)
3. Step-By-Step Initial-Valued Problem.
Use separation of variables to solve the initial value problem
\begin{equation*}
\frac{du}{dy} - u^2y = u^2, \hspace{1cm} u(0) = 5\text{.}
\end{equation*}
Select the variable that you are solving for: \(\quad\) \(\, y \,\) \(\quad\) \(\, u \,\)
Select the order of the equation: \(\quad\) \(\, 1 \,\) \(\quad\) \(\, 2 \,\) \(\quad\) \(\, 3 \,\) \(\quad\) \(\, 4 \,\)
Select the correct separable form:
\(\dfrac{du}{dy} = u^2(y-1) \) \(\quad\)
\(\dfrac{du}{dy} = u^2y \) \(\quad\)
\(\dfrac{du}{dy} = u^2(1+y) \) \(\quad\)
\(\dfrac{du}{dy} = -y \)
Select the next step following the correct answer to (c).
\(\dfrac{1}{y+1}dy = u^2du \) \(\quad\)
\(\dfrac{1}{u^2}du = (y+1)dy \) \(\quad\)
\(\dfrac{1}{y}dy = u^2 du \)
Select the result after integrating the correct answer to (d).
\(\ds\ln|u^2| = \frac{y^2}{2} + y + c \) \(\quad\)
\(\dfrac{1}{u} = \frac{y^2}{2} + y + c \) \(\quad\)
\(\ds-\frac{1}{u} = \frac{y^2}{2} + y + c \)
Select the correct general solution.
\(\ds u = \left(-\frac{y^2}{2} + y + c\right)^{-1} \) \(\quad\)
\(\ds u = c + \left(-\frac{y^2}{2} + y\right)^{-1} \) \(\quad\)
\(\ds u = ce^{\large y^2 + y} \) \(\quad\)
Select the correct particular solution.
\(\ds u = \left(-\frac{y^2}{2} + y + 5\right)^{-1} \) \(\quad\)
\(\ds u = \left(-\frac{y^2}{2} + y + \frac15\right)^{-1} \) \(\quad\)
\(\ds u = 5e^{\large y^2 + y } \)
4. \(\ds y'= e^{2x} - 4x\) .
Activate
\(_{}\)
\(\text{order:}\) \(\displaystyle \frac{dy}{dx} =\)
\(_{}\)
\(_{}\)
\(\displaystyle\int\) \(\displaystyle dy = \int\) \(dx\)
\(_{}\)
\(_{}\)
\(+ c_1 =\) \(+ c_2\)
\(_{}\)
5. \(\dfrac{dy}{dx} = xy^2\) .
Activate
\(_{}\)
\(\text{order:}\) \(\displaystyle \frac{dy}{dx} =\)
\(_{}\)
\(_{}\)
\(\displaystyle\int\) \(\displaystyle dy = \int\) \(dx\)
\(_{}\)
\(_{}\)
\(+ c_1 =\) \(+ c_2\)
\(_{}\)
6. \(\dfrac{dy}{dx} - y\cos(x) = y\) .
Activate
\(_{}\)
\(\text{order:}\) \(\displaystyle \frac{dy}{dx} =\)
\(_{}\)
\(_{}\)
\(\displaystyle\int\) \(\displaystyle dy = \int\) \(dx\)
\(_{}\)
\(_{}\)
\(+ c_1 =\) \(+ c_2\)
\(_{}\)
Level 1. Solve using separation of variables, if possible.
7. \(\dfrac{dy}{dx} = y^2\) 8. \(y^2 y' - 1 = 6x^2\) 9. \(\dfrac{dy}{dx} = x^2y\) 10. \(\dfrac{dy}{dx} = e^{x+y}\) 11. \(\dfrac{dy}{dx} = \frac{x}{y}\) 12. \(\dfrac{dy}{dx} = \cos(x) \sec(y)\) 13. \(\dfrac{dy}{dx} = \frac{2x}{1+y^2}\) 14. \(\dfrac{dy}{dx} = 2-y\) 15. \(\dfrac{dy}{dx} = \frac{y-x}{y+x}\) 16. \(\dfrac{dy}{dx} = y^2 \sec^2(x)\) 17. \(\dfrac{dy}{dx} = \frac{1}{xy}\)
Level 2. Solve using separation of variables, if possible.
18. \(\dfrac{dy}{dx} = x y\) 19. \(\dfrac{dy}{dx} = \dfrac{x^2 + 1}{y}\) 20. \(\dfrac{dy}{dx} = \dfrac{1 - x^2}{y^2}\) 21. \(y' - 2y = y\sin x\) 22. \(x\frac{dv}{dx} = \dfrac{1-4v^2}{3v}\) 23. \(\dfrac{dy}{dx} = \dfrac{4-2x}{3y^2-5}\) 24. \(\dfrac{dy}{dx} = 6x(y-1)^{2/3}\)
Initial Value Problems. Solve each of the following initial value problems using separation of variables, if possible.
25. \(\dfrac{dy}{dx} = -6xy;\ y(0)=7\) 26. \(\dfrac{dz}{dt} = 2tz^2;\ z(1) = 2\) 27. \(\dfrac{dy}{d\theta} = y\sin\theta;\ y(\pi) = -3\) 28. \(y' - 8x^3e^{-2y} = 0;\ y(1) = 0\)
Nested Differential Equations.
Find a non-zero function
\(\mu(x)\) that satisfies each of the following differential equations.
Hint .
For each equation, expand the derivative on the left side using the product rule. Then use algebra to find another differential equation nested within the original one.
29. \(\dfrac{d}{dx}\left[\mu\sin x\right] = \mu \cos x + 6 \mu \sin x\) 30. \(\dfrac{d}{dx}\left[\mu\ln x\right] = \dfrac{\mu}{x} + 2x \mu \ln x\) 31. \(\dfrac{d}{dx}\left[\mu y \right] = \mu \dfrac{dy}{dx} + 5 \mu y\quad \) (\(y\) is some function of \(x\) )
Applications. 32. \(50^{\circ}F\text{,}\) is placed in a \(375^{\circ}F\) oven at 5:00 PM. After 75 minutes, it is found that the temperature \(T(t)\) of the roast is \(125^{\circ}F\text{.}\) When will the roast be \(150^{\circ}F\) (medium-rare)?
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