Find the general solution to
$\displaystyle\frac{dy}{dx}=2xe^y$
Find the general solution to
$\displaystyle\frac{dy}{dx}=y(1-y)$
Find the general solution to
$\displaystyle x+\frac{dy}{dx}=xe^y$
Find the general solution to
$\displaystyle\frac{dy}{dx}+x^2y+y=0$
Solve the initial value problem
$\displaystyle\frac{dy}{dx}=y^2-9,$ $y(0)=1.$
Solve the initial value problem
$\displaystyle\frac{dy}{dx}=xy-3x,$ $y(0)=0.$
Solve the initial value problem
$\displaystyle\frac{dy}{dx}=\cos^2(y),$ $y(0)=0.$
Solve the initial value problem
$\displaystyle\frac{dy}{dx}=\cos^2(y),$ $y(0)=\pi/2.$
Find the family of orthogonal trajectories to the ellipses $x^2+4y^2=C^2$.
Find the family of orthogonal trajectories to the parabolas $y^2 + 2x = C$.
Find $y(1)$ where $y(x)$ is the solution to
$\displaystyle\frac{dy}{dx}=2y,$ $y(0)=1.$
Find $y(2)$ where $y(x)$ is the solution to
$\displaystyle\frac{dy}{dx}-y=xy,$ $y(0)=1.$
Find $y_0$ given that the solution to
$\displaystyle\frac{dy}{dx}=xy^2$, $y(0)=y_0,$ satisfies $y(1)=1.$
Find $y_0$ given that the solution to
$\displaystyle\frac{dy}{dx}=e^{x+y}$, $y(0)=y_0,$ satisfies $y(1)=2.$
Find $y_0$ so that the integral curve for
$$\begin{align}
&\displaystyle\frac{dy}{dx} = \displaystyle\frac{y}{2} -
\displaystyle\frac{x}{3} \\
&y(-4) = y_0
\end{align}$$ is
a straight-line. You must justify your answer, which will require you to
apply algebraic reasoning to the problem. Just using the slope field lab
applet to find the answer by repeated guessing will not get you full
credit.
Find $y_0$ so that the integral curve for
$$\begin{align}
&\displaystyle\frac{dy}{dx} = x+y \\
&y(-4) = y_0
\end{align}$$ is
a straight-line. You must justify your answer, which will require you to
apply algebraic reasoning to the problem. Just using the slope field lab
applet to find the answer by repeated guessing will not get you full
credit.
Find an initial value problem that is solved by
$y = e^{x^2}$.
Find an initial value problem that is solved by
$y = e^{\sin(x)}$.
Find a differential equation whose general solution is
the set of curves, $e^{x}+e^{y}=C^2.$ Remember
that $C$ is the arbitrary constant here so it should not appear in
the differential equation.
Find a differential equation whose general solution is
the collection of all circles about the origin, $x^2+y^2=r^2$. Remember
that $r$ is the arbitrary constant here so it should not appear in
the differential equation.
Technical note: Since a circle is not the graph of a
function as it fails the vertical line test, it would be more proper to
say the general solution consists of all semicircles of the form
$y=\pm\sqrt{r^2-x^2}$. But the problem is easier to understand and to
solve written as above.
Classify whether the following equations are exact,
linear, both, or neither. You don't have to solve them, just classify
them.
$(x+y)dx + dy = 0$
$(x^2+ye^x)dx + (e^x-y)dy = 0$
$(1+y)dx + dy = 0$
$(x+y^2)dx + (x^2+y)dy = 0$
Find the general solution to
$\displaystyle\frac{dy}{dx}=\exp(x)-\frac{y}{x}.$
Find the general solution to
$\displaystyle\frac{dy}{dx}=\frac{\exp(x)-y}{x-\sin(y)}.$
Find the general solution to $(x+y)dx+dy=0.$
Find the general solution to $(2x+y)dx+(x-2y)dy=0.$
Solve the initial value problem
$\displaystyle\frac{dy}{dx}+2y=\exp(x)$, $y(0)=2.$
Solve the initial value problem
$\displaystyle\frac{dy}{dx}-3y=x-1$, $y(0)=0.$
Solve the initial value problem
$(x^2+2xy)dx+(x^2-y)dy=0$, $y(0)=1.$
Solve the initial value problem
$(\cos(x)-y^2)dx+(\cos(y)-2xy)dy=0$, $y(0)=0.$
Suppose we have an ideal two tank CSTR (continuous stirred-tank reactor) in series. Fluid enters the first tank
and then flows from the first tank to the second tank and then out the second tank. We suppose the tanks each hold
2000 liters and the fluid is flowing at a rate of 10 liters/minute. Suppose a contaminant is introduced to
the fluid prior to entering the first tank, at a concentration of 2 mg/liter. What is the concentration of the
contaminant in each tank after 120 minutes? (Note: since this is an "ideal" model, we assume perfect mixing and
steady flow).
Suppose we have an ideal two tank CSTR (continuous stirred-tank reactor) in series. Fluid enters the first tank
and then flows from the first tank to the second tank and then out the second tank. We suppose the tanks each hold
600 liters and the fluid is flowing at a rate of 3 liters/minute. Suppose a contaminant is introduced to
the fluid prior to entering the first tank, at a concentration of 4 mg/liter. What is the concentration of the
contaminant in each tank after $t$ minutes? (Note: since this is an "ideal" model, we assume perfect mixing and
steady flow).
Suppose we have an ideal two tank CSTR (continuous stirred-tank reactor) in series. Fluid enters the first tank
and then flows from the first tank to the second tank and then out the second tank. We suppose the tanks each hold
2000 liters and the fluid is flowing at a rate of 4 liters/minute. Suppose 100g of a contaminant is accidentally introduced
to the first tank all at once (with no additional contamination after this). Then the maximum concentration of contaminant
in the first tank is 100 grams/2000 liters = 0.05 g/l = 50mg/l. What is the maximum concentration of contaminant in the second
tank and when does it occur?
Suppose we have an ideal two tank CSTR (continuous stirred-tank reactor) in series. Fluid enters the first tank
and then flows from the first tank to the second tank and then out the second tank. We suppose the tanks each hold
4000 liters and the fluid is flowing at a rate of 10 liters/minute. Suppose 250g of a contaminant is accidentally introduced
to the first tank all at once (with no additional contamination after this). Then the maximum concentration of contaminant
in the first tank is 250 grams/4000 liters = 0.0625 g/l = 62.5mg/l. What is the maximum concentration of contaminant in the second
tank and when does it occur?
Find $y(2)$ where $y(x)$ is the solution to the initial value problem
$$
\begin{align}
\frac{dy}{dx}&=\frac{x-y}{x+2y} \\
y(0)&=1
\end{align}
$$
Note that solving this exact equation will lead to an implicit formula for
$y$ which will have two solutions for $x=2$. However, the initial value
problem has a unique solution and so just one of the two solutions
produced from the implicit formula will be the actual value of $y(2).$
Justify your choice of value for $y(2).$ You can do this algebraically
(say by solving the implicit formula to get an explicit formula for $y$ in
terms of $x$ where the $\pm$ term will be specified to either be $+$ or
$-$) or geometrically (perhaps by arguing from the slope field that the
integral curve starting at (0,1) can only reach one of the two possible
values because it can't go below the x-axis). Just saying you checked the
slope field applet and can see
which value must be right won't get full credit, you must justify how
you can tell the integral curve can't go below the x-axis.
Find $y(1)$ where $y(x)$ is the solution to the initial value problem
$$
\begin{align}
\frac{dy}{dx}&=\frac{2x-2y}{2x+y} \\
y(0)&=\sqrt{3}
\end{align}
$$
Note that solving this exact equation will lead to an implicit formula for
$y$ which will have two solutions for $x=1$. However, the initial value
problem has a unique solution and so just one of the two solutions
produced from the implicit formula will be the actual value of $y(1).$
Justify your choice of value for $y(1).$ You can do this algebraically
(say by solving the implicit formula to get an explicit formula for $y$ in
terms of $x$ where the $\pm$ term will be specified to either be $+$ or
$-$) or geometrically (perhaps by arguing from the slope field that the
integral curve starting at (0,1) can only reach one of the two possible
values because it can't go below the x-axis). Just saying you checked the
slope field applet and can see
which value must be right won't get full credit, you must justify how
you can tell the integral curve can't go below the x-axis.
Solve the initial value problem
$$
\begin{align}
\frac{dy}{dx} &= 1-\cos(x)y \\
y(0)&=2
\end{align}
$$
and then use your solution to evaluate $y(2).$
Your answer should be accurate to the nearest 0.0001.
Note that you will get an integral here that you can't compute
analytically. You will need to use a numerical method to complete the
problem (the TI-83/84 and
similar calculators have an excellent numerical integration routine built
in, or you can use
WolframAlpha.)
Solve the initial value problem
$$
\begin{align}
\frac{dy}{dx}&=-\frac{y+\exp(x)}{x+\cos(y)} \\
y(0)&=0
\end{align}
$$
and then use your solution to evaluate $y(2).$
Your answer should be accurate to the nearest 0.0001.
In this case you will get an implicit formula that you won't be able to
solve analytically for $y$ in terms of $x$. You can substitute $2$ for $x$
in the solution to get an equation for $y$ which you can then solve
numerically quickly using Newton's method (which you can find described in
your calculus textbook or
online). Since
Newton's method has better convergence properties with a good
starting value, you may want to approximate $y(2)$ using Euler's method or
the improved Euler's method with a step-size of 1 or 2 to get your initial
guess. Note that the work to finish the solution using Newton's method is
much less than the work to get a numerical approximation to such a fine
tolerance using the numerical methods like Euler's method to approximate the solution
directly.
Solve the initial value problem
$$
\begin{align}
\frac{dy}{dx}&=\frac{\sin(x)-\exp(x+y)+2xy}{3y^2-x^2+\exp(x+y)} \\
y(0)&=0
\end{align}
$$
and then use your solution to evaluate $y(1).$
Your answer should be accurate to the nearest 0.0001.
In this case you will get an implicit formula that you won't be able to
solve analytically for $y$ in terms of $x$. You can substitute $1$ for $x$
in the solution to get an equation for $y$ which you can then solve
numerically quickly using Newton's method (which you can find described in
your calculus textbook or
online). Since
Newton's method has better convergence properties with a good
starting value, you may want to approximate $y(1)$ using Euler's method or
the improved Euler's method (a step-size of 1 will be fine) to get your
initial
guess. Note that the work to finish the solution using Newton's method is
much less than the work to get a numerical approximation to such a fine
tolerance using the numerical methods like Euler's method to approximate the
solution directly.
Find the general solution for $\exp(y)dx + (2x\exp(y)-1)dy = 0$. Hint: if
you treat
$x$ as a function of $y$, the equation is linear, i.e. it has an
integrating factor that is a function of $y$ alone.
Find the general solution for $-y\,dx + (x+x^2+y^2)dy = 0$. Hint:
$\displaystyle\frac{1}{x^2+y^2}$ is an integrating factor.
Show that the equation $(x+4y)dx + (3x+6y)dy = 0$ has an integrating
factor of the form $(x+y)^r$ for some value of $r.$ This will be
true for any equation of the form $(ax+by)dx + ((a+c)x+(b+c)y)dy = 0,$
where the value of $r$ will depend on the values of $a$, $b$, and $c$, but
you don't have to justify the general case.
Suppose $w(x)$ solves the initial value problem
$$
\begin{align}
\frac{dw}{dx} + p(x)w &= q(x) \\
w(0)&=0
\end{align}
$$
while $z(x)$ solves the initial value problem
$$
\begin{align}
\frac{dz}{dx} + p(x)z &= 0 \\
z(0)&=y_0
\end{align}
$$
Show that $y(x)=w(x)+z(x)$ solves the initial value problem
$$
\begin{align}
\frac{dy}{dx} + p(x)y &= q(x) \\
y(0)&=y_0
\end{align}
$$
The ability to cut a problem into pieces, solve each piece separately, and
add up the answers to each piece to find an answer to the overall problem
is a fundamental feature of "linear" equations in mathematics. We will
discuss this more in chapter 2.
Suppose you are modeling a situation governed by the equation
$$\frac{dx}{dt}+ax=f(t),\qquad x(0)=x_0$$ with $a>0$,
but due to measurement errors, your measured values $\tilde{f}(t)$ differ from the true
values $f(t)$. Let $\tilde{x}(t)$ be the solution to the equation based on the imperfect data
$$\frac{d\tilde{x}}{dt}+a\tilde{x}(t)=\tilde{f}(t),\qquad \tilde{x}(0)=x_0.$$
Show that the difference between the true solution and the imperfect solution, $w(t)=x(t)-\tilde{x}$,
is the solution to the initial value problem
$$\frac{dw}{dt}+aw=f(t)-\tilde{f}(t),\qquad w(0)=0.$$
Show that in the previous problem, if the error in your measurement is bounded by $b$, i.e.
$|f(t)-\tilde{f}(t)|\le b$ for all $x>0$, then the error in your solution will be bounded by
$b/a$, i.e. $|x(t)-\tilde{x}(t)|\le b/a$ for all $x>0$. Hint: In the previous problem you showed $x(t)-\tilde{x}(t)$ satisfied a linear initial value problem,
so you can write the formula for the solution of a linear differential equation will give you an
integral equal to $x(t)-\tilde{x}(t)$.
Then use the fact that $|\int(g(t)\,dt|\le \int|g(t)|\,dt$ and the bound on $|f(t)-\tilde{f}(t)|$ to get the required
bound on the error in $\tilde{x}(t)$.
Find the general solution to
$\displaystyle\frac{dy}{dx}+2y=\exp(x)y^2.$
Find the general solution to
$\displaystyle\frac{dy}{dx}=\frac{3x+13y}{15x+y}.$
Solve the initial values problem
$\displaystyle\frac{dy}{dx}=\frac{3y-2x}{4y-6x}$, $y(1)=2.$
Solve the initial values problem
$\displaystyle\frac{dy}{dx}-3y=\frac{x}{y}$, $y(0)=1.$
In the Change of Variables section we went over two types of equations
that can be solved by making substitutions, Bernoulli equations and
homogeneous equations. There are a lot of other substitutions that apply
to other types of equations. The keys for all the different substiution
methods are to rewrite the original variable $y$ in terms of the new
variable $v$ and to be sure to make the substitution everywhere in the
equation, including making the substitution in the derivative term, which
may require use of derivative rules such as the chain rule (as in
Bernoulli equations) or the product rule (as in homogeneous equations).
Two other common types of equations that can be solved by substitutions are
Any equation of the form $\displaystyle\frac{dy}{dx}=f(\alpha x+\beta
y+\gamma)$, where $\alpha$, $\beta$, and $\gamma$ are constants, can be
transformed to a separable equation via the substitution $v=\alpha x +
\beta y +\gamma$.
Equations of the form $\displaystyle\frac{dy}{dx}=A(x)y^2+B(x)y+C(x)$
are called Riccati equations. If you know any one particular solution,
$y_1(x)$, to a Riccati equation, you can find the general solution by
transforming the Riccati equation to a linear equation with the
substitution $v=\displaystyle\frac{1}{y-y_1(x)}$.
Using these tricks you can solve problems 59-64 below. Of course, to solve
a Riccati equation you must first find one particular solution, but you can
find one by guess and check more often than you might imagine. In context,
you may know one obvious solution, or you might look at the slope field
and be able to spot one simple solution like $y=1$ or $y=x$ or
$y=\exp(x)$ (especially if the problem appears in an undergraduate math
class).
Find the general solution to
$\displaystyle\frac{dy}{dx}=\frac{x-1+y-xy^2}{x}.$
Find the general solution to
$\displaystyle\frac{dy}{dx}=y^2-2xy+x^2.$
Find the general solution to
$\displaystyle\frac{dy}{dx}=2\exp(x)-\exp(-x)y^2.$
Solve the initial value problem
$\displaystyle\frac{dy}{dx}=\exp(x-y)+1,$
$y(0)=1$.
Solve the initial value problem
$\displaystyle\frac{dy}{dx}=y^2-2xy+x^2+1,$
$y(0)=1$.
Solve the initial value problem
$\displaystyle\frac{dy}{dx}=(x^2-x+1)+(1-2x)y+y^2,$
$y(0)=1.$
Find and classify (as stable, unstable, or semi-stable) all the
equilibrium solutions to
$\displaystyle\frac{dy}{dx}=y^3-4y^2+3y$
$\displaystyle\frac{dy}{dx}=(y^2+4y+3)(y^2-1)$
Find and classify (as stable, unstable, or semi-stable) all the
equilibrium solutions to
$\displaystyle\frac{dy}{dx}=\sin(y)$
$\displaystyle\frac{dy}{dx}=\cos(y)$
Suppose $\displaystyle\frac{dy}{dx}=x^2-y^2$, $y(1)=0.$ Explain how we can
tell that $0 < y(x) < x$ for all $x > 1$. Hint: sketch the slope
field, specifically noting how the slope field looks along the lines $y =
x$ and $y = 0$. Explain why an integral curve
can't leave the wedge in the first quadrant formed by these two lines.
Two students are asked if the solution to the initial value problem.
$\displaystyle \frac{dy}{dx}=3x^2y^2$, $y(0)=1$ can ever take a negative
value for $x > 0.$ Kasey argues that $y$ can't be negative because
$3x^2y^2 > 0,$ so the derivative is
always non-negative, hence $y$ is always increasing or flat, and, since we
increase from $y = 1,$ $y$ must always be positive. On the other hand,
Riley solves the problem to get
$\displaystyle y = \frac{1}{1-x^3}$ and computes
$\displaystyle y(2) = \frac{1}{1-2^3} = -\frac17 < 0$
hence y can be negative. Write a paragraph explaining whether Kasey or
Riley (or both or neither) is correct.
Suppose a fish population experiences logistic growth with growth rate $r$
per
year and environmental carrying capacity $C$. The population is harvested
at a rate of $h$ per year ($h$ might be a constant or it might be
variable depending on the population). This situation is described by the
differential equations
$$
\frac{dp}{dt}=rp(1-p/C)-h
$$
where $p$ is the population and $t$ is time (in years).
It is desired to harvest the
fish to reach the maximum sustainable yield. That is, if you harvest too
few fish, then you aren't getting all the fish you could. But if you
harvest too many fish, you will drive the population too low (perhaps to
extinction) and so your harvest won't be sustainable. The next 4 problems
take you through the calculations of the maximum sustainable yield for two
simple harvesting models.
Suppose the growth rate is $r=0.2$ (20%), the carrying capacity is
$C=25,000,$ and the harvest is a constant $H$ fish per year.
What are the equilibria for this model? Are they stable or
unstable?
What is the maximum value of $H$ so that the model has an equilibrium
with $p > 0$?
Hint: as always when determining equilibria, drawing a graph of $dp/dt$
vs. $p$ may be helpful.
Suppose the harvest is a constant $H$ fish per year. What is the
maximum value of $H$ so that the model has an equilibrium with $p > 0$?
Your answer this time will have $H$ as a function of $r$ and $c$, in
contrast to the previous problem where you were given specific values for
$r$ and $C.$
Suppose the harvest is not a constant but is a fixed proportion of the
total population, so $h=Ep$ where $E$ is the proportion of the population
that is harvested.
What are the equilibria for the model now?
Which equilibrium is stable and which is unstable?
Note that the equilibrium populations will depend on the value of $E$, so
your answers will be functions $E$ as well as $r$ and $C$
Assuming you end up at the stable equilibrium in problem 67, what
proportion $E$ maximizes the harvest (recall $h=Ep$ where $p$ is now a
function of $E$, $r$, and $C$)? What is the maximum sustainable
yield $h$ for this approach?
Many policies were originally designed to aim for the maximum
sustainable yield as computed above. However, this
model is quite simple and the
target values developed using it sometimes turned out to be too high in practice.
Find the general solution to $x(y')^2-yy' + 1=0$. You don't need to find the singular solution.
(Hint: If you solve for $y$ in terms of $x$ and $y'$, this is a Clairaut equation).
Find the general solution to $(2x+1)(y')^2 - 2(y+2)y' + 4 = 0$. You don't need to find the singular solution.
(Hint: If you solve for $y$ in terms of $x$ and $y'$, this is a Clairaut equation).
Find a Clairaut differential equation whose general solution is all the tangent lines to $y=x^2+1$.
Find a Clairaut differential equation whose general solution is all the tangent lines to $y=e^x$.
In the section on "Theory of First-order Equations", the Clairaut equation $y = xy' - (y')^2/4$
was solved, and it was pointed out that you could stitch different solutions together to get
new solutions. Consider the function
$$h(x)=\begin{cases}
-2x-1 \text{ for }x<0, \\
2x-1\qquad\text{ for }x\ge 0.
\end{cases}$$
Is this a solution to $y=xy'-(y')^2/4$? Explain why or why not.
Find a solution to the Clairaut equation $y=xy' - (y')^2/4$ which passes through the points $(-2,5)$ and $(3,7)$. Hint:
You will have to stitch several basic solutions together.
An envelope of a family of curves is a curve that is tangent to each member of the family at some point and these
points of tangency form the whole enveloping curve (note that most families of curves do not have an enveloping curve).
Suppose a family of curves all solve a first-order differential equation, $F(x,y,y')=0$. Show that if the family of curves
has an enveloping curve, then that envelope also solves the differential equation.
Consider the initial value problem $\displaystyle \frac{dy}{dx}=2x(y^2+1)$, $y(0)=0$. the true solution
is $y(x)=\tan(x^2)$, but here we will see how Picard iteration works for this problem. The iteration
will start with $y_0(x)=0$. Compute
$y_2(x)$ (this time you have to work out the integral to evaluate for yourself).
Consider the initial value problem $\displaystyle \frac{dy}{dx}=y$,
$y(0)=1.$ The true solution is $y(x)=\exp(x),$ but here we will see how
Picard iteration works for this problem. The iteration will start with
$y_0(x)=1.$ Compute
$\displaystyle y_1(x) = \int_0^x y_0(t)dt + 1$
$y_2(x)$ (this time you have to work out the integral to evaluate for
yourself).