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Section Chapter 2 Exercises
Reading Questions α―β
β Quick-Answer Questions
1. True-False.
(a) True or False .
In general, differential equations have more than one solution.
True.
True. In general, differential equations have infinitely-many solutions.
False.
True. In general, differential equations have infinitely-many solutions.
(c) True or False .
The differential equation
\begin{equation*}
\frac{dy}{dx} = 2xy - 6x\text{,}
\end{equation*}
is an initial value problem.
True.
False. This differential equation is not an initial value problem because it does not include any known values of the solution. An example of an initial value problem would be
\begin{equation*}
\frac{dy}{dx} = 2xy - 6x, \hspace{0.25cm} y(5) = -10\text{,}
\end{equation*}
since it specifies that \(y\) must be \(-10\) when \(x = 5\text{.}\)
False.
False. This differential equation is not an initial value problem because it does not include any known values of the solution. An example of an initial value problem would be
\begin{equation*}
\frac{dy}{dx} = 2xy - 6x, \hspace{0.25cm} y(5) = -10\text{,}
\end{equation*}
since it specifies that \(y\) must be \(-10\) when \(x = 5\text{.}\)
(e) True-or-False .
\(y = x^2 + 3\) is a solution to the differential equation
\begin{equation*}
\frac{dy}{dx} - 3 = 2x\text{.}
\end{equation*}
Incorrect. \(y = x^2 + 3\) is not a solution since
\begin{align*}
\frac{dy}{dx} - 3 \amp = 2x \\
\frac{d}{dx}\left[x^2 + 3\right] - 3 \amp = 2x \\
2x - 3 \amp = 2x \quad \leftarrow \text{false}
\end{align*}
Correct! \(y = x^2 + 3\) is not a solution since
\begin{align*}
\frac{dy}{dx} - 3 \amp = 2x \\
\frac{d}{dx}\left[x^2 + 3\right] - 3 \amp = 2x \\
2x - 3 \amp = 2x \quad \leftarrow \text{false}
\end{align*}
(f) True-or-False .
A family of solutions can be viewed as the collection of all particular solutions.
Correct! A family of solutions includes all possible particular solutions.
Incorrect. A family of solutions is a set of all possible solutions, not just one particular solution.
(g) True-or-False .
A particular solution can be viewed as a member of the family of solutions.
(i) True-or-False .
A family of solutions is the collection of all general solutions.
True.
This statement is false . A family of solutions includes all particular solutions, which are derived from the general solution by assigning specific values to the constants. The general solution itself is a single form that represents many solutions, not a collection of them.
False.
This statement is false . A family of solutions includes all particular solutions, which are derived from the general solution by assigning specific values to the constants. The general solution itself is a single form that represents many solutions, not a collection of them.
2. Short-Answer Questions.
(a) Algebraic vs. Differential Equation Solutions.
Consider the algebraic equation
\begin{equation*}
2x^2 + 3 = 7x\text{.}
\end{equation*}
State what it means for
\(x = 3\) to be a solution to this equation.
Show how you would verify that
\(x = 3\) is a solution.
Do not solve for \(x\text{.}\)
Show how you would verify that
\(x = 4\) is a solution.
Are there any differences in how you verify solutions to differentials equations compared to algebraic equations? Explain.
(b) Explain the Difference.. .
In few sentences, explain the difference between a general solution, family of solutions, and a particualr solution.
(c) Explain the Significance.. .
Explain the significance of initial condition(s) as they relate to the particular and general solution of a differential equation.
(d)
Attempt to apply direct integration to the differential equation
\begin{equation*}
\frac{dy}{dx} = x + y\text{.}
\end{equation*}
Get to the point where it becomes clear that you cannot solve for \(y\) directly. What is the obstacle?
Solution 1 . (a)
To be a solution to the equation,
\(x = 3\) must satisfy the equation. That is, when we substitute
\(x = 3\) into the equation, the result simplifies to a true statement.
Substituting \(x = 3\) into the equation, we get
\begin{align*}
\os{\text{LHS}}{\overline{2x^2 + 3}} \amp = \os{\text{RHS}}{\overline{7x}} \\
2(3)^2 + 3 \amp = 7(3) \\
21 \amp = 21 \quadβ
\end{align*}
Since \(21=21\) is an undeniably true statement, we have confirmed that \(x = 3\) is a solution to the equation.
Substituting \(x = 4\) into the equation, we get
\begin{align*}
\os{\text{LHS}}{\overline{2x^2 + 3}} \amp = \os{\text{RHS}}{\overline{7x}} \\
2(4)^2 + 3 \amp = 7(4) \\
35 \amp = 28 \quadβ
\end{align*}
Since \(35\ne28\) is not true, so \(x = 4\) is not a solution to the equation.
The process of verifying solutions to differential equations is exactly the same. However, with differential equations, we have to be more careful about confirming a βtrue statementβ. With numbers, it is easy to see if \(21=21\) is true, but with functions, its a bit more tricky. We have to be sure that the functions are equal for all values of \(x\) (or the independent variable). For example, the statement
\begin{equation*}
e^x + \sin^2 x + \cos^2 x = e^x + 1
\end{equation*}
is true since \(\sin^2 x + \cos^2 x = 1\text{.}\) In contrast, the statement
\begin{equation*}
e^x + \sin x + \cos x = e^x + 1
\end{equation*}
is not true since \(\sin x + \cos x \ne 1\) for all values of \(x\text{.}\) It only takes one value of \(x\) to make the statement false. Letβs try a few values of \(x\) to see this.
\begin{equation*}
x = 0:
\end{equation*}
\begin{align*}
e^0 + \sin 0 + \cos 0 \amp = 1 + 1 \\
2 \amp = 2 \quadβ
\end{align*}
\begin{equation*}
x = \frac{\pi}{2}:
\end{equation*}
\begin{align*}
e^{\pi/2} + \sin \frac{\pi}{2} + \cos \frac{\pi}{2} \amp = e^{\pi/2} + 1 \\
e^{\pi/2} + 1 \amp = e^{\pi/2} + 1 \quadβ
\end{align*}
\begin{equation*}
x = \pi:
\end{equation*}
\begin{align*}
e^{\pi} + \sin \pi + \cos \pi \amp = e^{\pi} + 1 \\
e^{\pi} - 1 \ne\amp\ e^{\pi} + 1 \quadβ
\end{align*}
The fact that the statement is not true for
\(x = \pi\) is enough to show that this is not a true statement and would not correspond to a solution to a differential equation.
Solution 2 . (b)
A general solution represents the form of all possible solutions to a differential equation, typically with one or more arbitrary constants. A family of solutions is an infinite set of solutions, one for each possible combination of constant values in the general solution. A particular solution is a single solution that satisfies the differential equation with specific values for the constants.
Solution 3 . (c)
The initial condition(s) specify one or more points that the graph of the solution must pass through. This often allows one to solve for the constants in the general solution. Therefore, the initial condition(s) act to reduce the family of solutions down to a smaller set of solutions or, ideally, a single particular solution.
Solution 4 . (d)
Integrating both sides gives
\begin{align*}
\int \frac{dy}{dx}\ dx \amp = \int\left(x + y\right)\ dx \\
y + C_1 \amp = \int x\ dx + \int y\ dx \\
y + C_1 \amp = \frac12 x^2 + C_2 + \int y\ dx \\
y - \int y\ dx \amp = \frac12 x^2 + C_2 - C_1
\end{align*}
Without knowing \(y\text{,}\) we cannot simplify \(\int y\ dx\text{.}\) So the obstacle is that we are unable to combine these \(y\) variables into a single \(y\) on the left side.
3. Solutions.
(a) Select-the-Best-Answer .
In general, what is a βsolutionβ to a differential equation?
function of the dependent variable.
While the solution is a function, it is not a function of the dependent variable.
function of the independent variable.
Yes, when you solve a differential equation, you are finding a function of the independent variable.
It is possible for a solution to be a number, but not in general.
derivatives of the dependent variable.
While derivatives of the dependent variable are involved, they do not describe solutions in general.
(b) Drag-and-Drop-Answer .
(c) Select-the-Best-Answer .
Consider the differential equation with missing right-hand side:
\begin{equation*}
y'' - \frac{4}{x}y' = \fillinmath{XXXXX}\text{.}
\end{equation*}
Assuming \(y = 2x^3\) is a solution to this equation, which of the following is a possible right-hand side?
(d) Select-the-Best-Answer .
What is a family of solutions?
A collection of all possible solutions to a differential equation.
Correct! The family of solutions includes every possible particular solution.
The general solution to a differential equation.
Incorrect. The general solution represents a form of the family of solutions, not the entire set.
A single specific solution to a differential equation.
Incorrect. This describes a particular solution, not the family of solutions.
A solution without any constants.
Incorrect. A solution without constants is typically a particular solution, not the entire family.
(e) Select-the-Best-Answer .
(f) Select-the-Best-Answer .
(g) Select-the-Best-Answer .
What does it mean to βsolveβ a differential equation?
To find an unknown function that satisfies the equation.
Correct! The goal of solving a differential equation is to find the function that meets the equationβs conditions.
To find the derivative of a function.
Incorrect. While derivatives are involved, the goal is to find the function, not just its derivative.
To identify the constants in an equation.
Incorrect. Identifying constants might be part of the process, but it is not the primary goal.
To determine the independent variable.
Incorrect. The independent variable is usually known; we solve for the dependent variable.
(h) Select-the-Best-Answer .
Which task is fundamentally different from the others?
Solving a differential equation.
Incorrect. Solving a differential equation is very similar to two other tasks in this list.
Finding the general solution to a differential equation.
Incorrect. Finding the general solution is very similar to two other tasks in this list.
Finding a family of solutions to a differential equation.
Incorrect. Finding a family of solutions is very similar to two other tasks in this list.
Verifying a solution to a differential equation.
Correct! Verifying a solution is much different than the other tasks that seek to find a solution.
(i) Select-the-Best-Answer .
Which statement is the best visual description of a particular solution to a differential equation?
The graph of a solution that passes through one or more specific points.
Correct! A particular solution is a single curve that satisfies the differential equation with a specific constant value, distinguishing it from the general solution.
The graph of multiple solutions to the equation.
Incorrect. This description fits a general solution, which encompasses all particular solutions by varying the parameter.
The intersection of all possible solutions to the equation.
Incorrect. This is not a visual representation of a particular solution.
Incorrect. The color of the curve does not define a particular solution.
5. Antiderivatives & Direct Integration.
(a) True-or-False .
We can solve
\begin{equation*}
\frac{dy}{dx} = x^3 - 7
\end{equation*}
for \(y\) by differentiating both sides with respect to \(x\text{.}\)
Incorrect, taking a derivative of both sides will result in a second derivative on the left side of the equation.
Correct! We should integrate both sides to solve for \(y\text{,}\) not differentiate.
(b) True-or-False .
Solving for \(y\) in the equation
\begin{equation*}
\frac{dy}{dx} = \ln(3x+1)
\end{equation*}
amounts to finding the antiderivative of \(\ln(3x+1)\text{.}\)
Correct, integrating both sides gives
\begin{equation*}
y = \int \ln(3x+1)\ dx \quad \leftarrow \text{anti-derivative of } \ln(3x+1)\text{.}
\end{equation*}
Incorrect.
(c) True-or-False .
Combining constants is a common practice in differential equations.
(d) Select-the-Best-Answer .
How could you solve for \(y\) in the equation
\begin{equation*}
\frac12\frac{dy}{dx} - \tan(2x) = x\text{?}
\end{equation*}
Differentiating both sides with respect to
\(x\text{.}\)
Incorrect, differentiating both sides only puts another derivative on \(\frac{dy}{dx}\text{.}\)
Isolate
\(\frac{dy}{dx}\) and integrating both sides with respect to
\(x\text{.}\)
Correct!
Isolate
\(\frac{dy}{dx}\) and integrating both sides with respect to
\(y\text{.}\)
Incorrect, the integration is not with respect to \(y\text{.}\)
Find the antiderivative of
\(\tan(2x)\text{.}\)
Incorrect, the solution is the antiderivative of \(2\tan(2x) + 2x\text{,}\) not just \(\tan(2x)\text{.}\)
(e) Select-the-Best-Answer .
The solution to the differential equation
\begin{equation*}
\frac13 y'- 7x + x^2 = 1
\end{equation*}
is the antiderivative of which function?
Incorrect. \(y\) is the solution to the differential equation.
Incorrect, perhaps check your algebra.
Incorrect, perhaps check your algebra.
Correct! Isolating \(y'\) gives
\begin{equation*}
y' = 21x - 3x^2 + 3\text{,}
\end{equation*}
so the solution is the antiderivative of \(21x - 3x^2 + 3\text{.}\)
(f) True-or-False .
Solving a differential equation by direct integration involves computing a derivative.
(g) True-or-False .
Direct integration could be used to solve the equation
\begin{equation*}
\frac{d}{dx}\left[y^2 + x^3\right] = \sqrt{x}\text{.}
\end{equation*}
Correct!
Incorrect. This equation is in the form
(8) .
(h) Select-the-Best-Answer .
Give the reason direct integration cannot be applied to the equation
\begin{equation*}
\frac{d}{dx}\left[\frac{x}{y^2}\right] = \sin(x+y)\text{.}
\end{equation*}
There is a fraction in the derivative.
Incorrect, direct integration doesnβt care about fractions.
The
\(y\) term is squared.
Incorrect, direct integration can handle this.
There is a sine term on the right side of the equation.
Incorrect, the sine is not the issue here.
The right-hand side contains
\(y\text{.}\)
Correct! Direct integration only works when the right-hand side contains only the independent variable, in this case \(x\text{.}\)
(i) Select-the-Best-Answer .
In the differential equation
\begin{equation*}
\frac{d}{dx}\left[5x \cdot y\right] = \frac{1}{x^2}\text{,}
\end{equation*}
what is the first step in solving for \(y\text{?}\)
Release
\(y\) by integrating both sides with respect to
\(x\text{.}\)
Correct! Integrating both sides is the first step in solving for \(y\text{.}\)
Release
\(x\) and
\(y\) by integrating both sides with respect to
\(y\text{.}\)
Incorrect. Integrating both sides with respect to \(y\) would not eliminate the derivative since the derivative is with respect to \(x\text{.}\)
Compute the derivative of
\(5x \cdot y\) using the product rule.
Incorrect. This would actually make the equation more complicated.
Incorrect. This would not help solve for \(y\text{.}\)
6. Fill in the Blanks.
The words, on the left, have been removed from the statements, on the right.
Drag each word to the panel that contains the statement it was removed from.
dependent
The solution is represented by the
general
particular
The
independent
The solution depends on the
satisfies
function
For a
7. Matching Each Function to the Equation it Satisfies.
Exercises ποΈ Warm-ups & Drills
Select the Solutions. For each differential equation, select the functions that are solution to that equation.
1. \(y''-9y = 0\) .
a. \(\ds \,\, y = 3\) b. \(\ds \,\, y = 0\) c. \(\ds \,\, y = e^{3x}\)
d. \(\ds \,\, y = 3x\) e. \(\ds \,\, y = 9e^x\) f. \(\ds \,\, y = x^9\)
g. \(\ds \,\, y = 4e^{3x}\) h. \(\ds \,\, y = e^{-3x}\) i. \(\ds \,\, y = e^{3x} - 2e^{-3x}\)
2. \(y'' - 10y' + 25y = 0\) .
a. \(\ds \,\, y = 0\) b. \(\ds \,\, y = x^2e^{5x}\) c. \(\ds \,\, y = e^{5x}\)
d. \(\ds \,\, y = 5x\) e. \(\ds \,\, y = xe^{5x}\) f. \(\ds \,\, y = 5e^{5x}\)
g. \(\ds \,\, y = 1/25\) h. \(\ds \,\, y = e^{-5x}\) i. \(\ds \,\, y = (1+x)e^{5x}\)
Find the Hidden Right-Hand Side. For each given
\(y(t)\text{,}\) assume it is a solution to the differential equation with hidden right side. Give the function that must be on the right for
\(y\) to be a solution to the equation.
3. \(\ y(t) = 2e^{-3t}\) .
4. \(\ y(t) = 3\sin(t^2)\) .
Exercises πΉοΈ Solutions to Differential Equations
Verifying Solutions. For each, verify if the given function,
\(y\text{,}\) is a solution to the differential equation.
1. \(y = c_1 \sin x + c_2 \cos x\text{,}\) \(\quad \dfrac{d^2y}{dx^2} + y = 0\) 2. \(y = c_{1}e^{2x} + c_{2}xe^{2x}\text{,}\) \(\quad \dfrac{d^{2}y}{dx^{2}} - 4\dfrac{dy}{dx} + 4y = 0\) 3. \(y = 3\sin(x^2)\text{,}\) \(\quad y' - xy'' = 12x^2\sin(x^2)\) 4. \(y = Ce^{-5x^2} + \dfrac{3}{5}\text{,}\) \(\quad y' = \big( 2x \big) \big( 3 - 5y \big)\)
Find the value of \(r\text{.}\) . Each of the following differential equations has one or more solutions of the form
\(y(t) = e^{rt}\text{,}\) for different values of constant,
\(r\text{.}\) Find all the values of
\(r\) so that
\(y\) is a solution to the equation.
5. \(\dfrac{d^2 y}{dx^2} + 6 \dfrac{dy}{dx} = -5y\) 6. \(y'' - 10y' + 24y = 0\) 7. \(y'' - 25y = 0\) 8. \(7y'' = 8y\)
Find the value of \(k\text{.}\) . Each of the following differential equations has one or more solutions of the form
\(y = t^{k}\ (t > 0)\text{,}\) for different values of constant,
\(k\text{.}\) Find all the values of
\(k\) so that
\(y\) is a solution to the equation.
9. \(3t^2 y^{\prime\prime} = -11t y^{\prime} + 3y\) 10. \(t^{2}y''- 11ty' + 32y = 0\)
Find the value of \(r\text{.}\) . Find the value of
\(r\) in
\(y\) or the differential equation such that
\(y\) is a solution to the equation.
11. \(y=rx^{2};\ y'=5x\) 12. \(y=\sin(rt);\ y'' + 9y = 0\) 13. \(y=e^{3t};\ \dfrac{d^2y}{dt^2} - 8 \dfrac{dy}{dt} + ry = 0\)
Solutions to Initial-Valued Problems.
14. Answer the following.
Show that \(y = \dfrac{\ln x + c}{x}\) is a solution to the differential equation
\begin{equation*}
x^2y' + xy = 1, \qquad y(9)=8\text{.}
\end{equation*}
Find
\(c\) corresponding to the initial condition.
15. Verifying Particular Solutions of Initial-Valued Problems Takes 2-Steps.
To verifying that a function (e.g.,
\(y = -3e^{x^2} + 3\) ) is a particular solution to an initial-value problem, there is an extra step beyond showing that it satisfies the differential equation. What else must this function satisfy?
16.
Show \(y = 3x^2 - 2x + 1\) is a particular solution to
\begin{equation*}
\frac{dy}{dx} = 6x - 2, \qquad y(0) = 1\text{.}
\end{equation*}
17.
Show \(y = -3e^{x^2} + 3\) is a particular solution to
\begin{equation*}
\frac{dy}{dx} = 2xy - 6x, \qquad y(0) = 0\text{.}
\end{equation*}
18. Find the Correct Solution.
Consider the differential equation
\begin{equation*}
y^\prime = y-y^2, \qquad y(0) = 7
\end{equation*}
and the following three functions
\begin{equation*}
y_1 = c\sin(-x), \quad y_2 = \frac{1}{c+x^2}, \quad y_3 = \frac{1}{1+ce^{-x}}\text{.}
\end{equation*}
Determine which, if any, of
\(y_1, y_2, \) or
\(y_3\) are solutions to this equation.
For any of the solutions found in (a), find the particular solution that corresponds to the initial condition.
Answer .
b.
\(\ds y=\frac{1}{1+ce^{-x}}\) \(c = \frac{1}{7}-1 \approx -0.857\text{,}\) so
\(\ds y = \frac{1}{1+(-0.857)e^{-x}} \) is the particular solution
Solution .
To show this, letβs first move all terms to one side of the equation, so
\begin{equation*}
\os{\large{\text{LHS}}}{\overline{y^{\prime}-y+y^2}} = 0
\end{equation*}
For each of the functions above, we will compute
\(y^\prime\) and plug it and
\(y\) into the left hand side (LHS) of the equation to see if it simplifies to zero.
\begin{equation*}
y=c\sin(-x) \quad \implies \quad y^\prime = -c\cos(-x)
\end{equation*}
\begin{align*}
\os{\large{\text{LHS}}}{\overline{y^{\prime}-y+y^2}}
\amp = -c\cos(-x) - c\sin(-x) + c^2\sin^2(-x)\\
\amp = -c\cos(x) + c\sin(x) + c^2\sin^2(x)
\end{align*}
It can be difficult to see why this doesnβt simplify to zero. However, if we assume it does, then plugging any \(x\) into this expression must also give you zero. So, if you can find one \(x\) value where this doesnβt happen, then this function cannot be a solution.
Letting
\(x=0\) gives
\(-c\text{,}\) but
\(c\) can be any constant, so
\(y = c\sin(-x)\) is not a solution.
\begin{equation*}
y = \frac{1}{c+x^2} \quad \implies \quad y^\prime = \frac{-2x}{(c+x^2)^2}
\end{equation*}
\begin{align*}
\os{\large{\text{LHS}}}{\overline{y^{\prime}-y+y^2}}
\amp = \frac{-2x}{(c+x^2)^2} - \frac{1}{c+x^2} - \frac{1}{(c+x^2)^2}\\
\amp = \frac{-2x}{(c+x^2)^2} - \frac{1}{c+x^2}\cdot \frac{c+x^2}{c+x^2} - \frac{1}{(c+x^2)^2}\\
\amp = \frac{-2x}{(c+x^2)^2} - \frac{c+x^2}{(c+x^2)^2} - \frac{1}{(c+x^2)^2} \\
\amp = \frac{-2x-c-x^2-1}{(c+x^2)^2} \ne 0 \quad β
\end{align*}
Letting
\(x=0\) gives
\(-1/c\text{,}\) but since
\(c\) can be any constant,
\(\ds y = \frac{1}{c+x^2}\) is not a solution.
\begin{equation*}
y = \frac{1}{1+ce^{-x}} \quad \implies \quad y^\prime = \frac{ce^{-x}}{(1+ce^{-x})^2}
\end{equation*}
\begin{align*}
\os{\large{\text{LHS}}}{\overline{y^{\prime}-y+y^2}}
\amp = \frac{ce^{-x}}{(1+ce^{-x})^2} - \frac{1}{1+ce^{-x}} - \frac{1}{(1+ce^{-x})^2}\\
\amp = \frac{ce^{-x}}{(1+ce^{-x})^2} - \frac{1+ce^{-x}}{(1+ce^{-x})^2} - \frac{1}{(1+ce^{-x})^2}\\
\amp = \frac{ce^{-x}-1-ce^{-x}-1}{(1+ce^{-x})^2} \\
\amp = \frac{0}{(1+ce^{-x})^2} = 0 \quad β
\end{align*}
So
\(\ds y = \frac{1}{1+ce^{-x}}\) is a solution.
Setting
\(y(0)=7\) we get
\begin{equation*}
7 = \frac{1}{1+c} \quad \implies \quad c = \frac{1}{7}-1 = -\frac{6}{7}
\end{equation*}
So the particular solution for this initial condition is
\begin{equation*}
y = \frac{1}{1-\frac{6}{7}e^{-x}}
\end{equation*}
Exercises βπ» Solving Differential Equations
General Solution. Find the general solution for each of the following differential equations. Combine constants where appropriate.
1. \(\ds \frac{dy}{dx} = 2x - 5\) 2. \(\ds \frac{d}{dx}[y] = e^{2x}\) 3. \(\ds \frac{d}{dx}[xy] = \cos x\) 4. \(\ds \frac{dy}{dt} = \frac{1}{t^2} + t\) 5. \(\ds \frac{dy}{dx} = x e^x\) 6. \(\ds \left[y\sin x\right]^\prime = \cos^2 x\) 7. \(\ds y' = \ln x + x^2\) 8. \(\ds \frac{d}{dP}\left[e^P Q\right] = P\)
Particular Solution. Find the particular solution for each of the following differential equations with the given initial condition.
9. \(\dfrac{dy}{dx} = 3x^2 + 2,\ y(1) = 4\) 10. \(\dfrac{d}{dt}[y] = \sin t,\ y(0) = 2\) 11. \(\dfrac{d}{dx}[x y] = e^x,\ y(1) = 0\) 12. \(\dfrac{dy}{dx} = x^3 - 4,\ y(2) = 1\) 13. \(\dfrac{dR}{dh} = \dfrac{1}{h} + h,\ R(1) = 3\) 14. \(y' = \cos x + x^2,\ y(0) = -2\) 15. \(\dfrac{d}{dx}\left[e^x y\right] = \sin x,\ y(0) = 2\) 16. \(2y' - 4\sin x = 2, \ y(0) = 5\) 17. \(\left[y\tan(x)\right]^{\prime} = \sec^2(x),\ y\left(\dfrac{\pi}{4}\right) = 1\)
18. Solve the Equation.
Solve the initial-value problem
\begin{equation*}
2y' - 4\sin x = 2, \quad y(0) = 5 \text{.}
\end{equation*}
Exercises Preview of a Future Method
At this point, you should be comfortable solving an equation such as
\begin{equation*}
\left[x^7 y \right]^{\prime} = e^x\text{.}
\end{equation*}
The problem is that most equations do not start in this form. Instead, they start in another form and then, after some algebra, are put into this nice form and solved. The process off rewriting an equation in this way is the basis for another technique called the integrating factor method. The question we want to answer here is βwhat type of equations can be written in this form?β
1. Give the equation that can be rewritten in the form \(\ds\left[x^7 y \right]^{\prime} = e^x\) .
Rewrite and Solve.
For each equation below, complete the following:
Use the product rule to rewrite each differential equation in the form
\begin{equation*}
y^{\prime} + P(x) y = Q(x)\text{.}
\end{equation*}
2. \(\ds \left[y \cos x \right]^{\prime} = \frac{1}{x^2}\) 3. \(\ds \left[y x^5 \right]^{\prime} = x\) 4. \(\ds \left[x^{-1}y\right]^{\prime} = e^{2x}\)
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