First Order ODEs Practice

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1.1Classify a Basic First-Order Equation

Exam I | Problem 1.1 |

For the equation

$$ y' = x^2 y, $$

state whether it is first-order and whether it is linear.

Solution

The highest derivative is $y'$, so the equation is first-order.

Rewrite it in standard linear form:

$$ y' - x^2 y = 0. $$

This matches

$$ y' + p(x)y = q(x), $$

with $p(x) = -x^2$ and $q(x)=0$. So the equation is linear.

1.2Read a Slope from the Differential Equation

Exam I | Problem 1.2 |

For

$$ y' = x - y, $$

what is the slope of the solution curve at the point $(3,1)$?

Solution

Substitute the point into the right-hand side:

$$ y' = 3 - 1 = 2. $$

So the slope at $(3,1)$ is $2$.

1.3Solve a Simple Separable Equation

Exam I | Problem 1.3 |

Separable Equations

Solve for $y$:

\frac{dy}{dx} = 4x.

Solution

Integrate both sides with respect to $x$:

y = \int 4x\,dx = 2x^2 + C.

So the general solution is

$$ y = 2x^2 + C. $$

1.4Apply an Initial Condition

Exam I | Problem 1.4 |

Initial Value Problems

Solve the initial value problem:

y' = 2x, \qquad y(1)=5.

Solution

Integrate the differential equation:

$$ y = x^2 + C. $$

Use the initial condition $y(1)=5$:

$$ 5 = 1 + C $$

$$ C = 4. $$

Therefore,

$$ y = x^2 + 4. $$

1.5Find Equilibrium Solutions

Exam I | Problem 1.5 |

For the autonomous equation

$$ y' = y(y-4), $$

find all equilibrium solutions.

Solution

Equilibria occur when the right-hand side is zero:

$$ y(y-4)=0. $$

y=0 \quad \text{or} \quad y=4.

The equilibrium solutions are $y(t)=0$ and $y(t)=4$.

1.6Test for Exactness

Exam I | Problem 1.6 |

Exact Equations

Determine whether

(2xy+1)\,dx + (x^2+3y^2)\,dy = 0

is exact.

Solution

Let

M(x,y)=2xy+1, \qquad N(x,y)=x^2+3y^2.

Compute the partial derivatives:

\frac{\partial M}{\partial y} = 2x, \qquad \frac{\partial N}{\partial x} = 2x.

Since these are equal, the equation is exact.

1.7Recognize a Homogeneous Equation

Exam I | Problem 1.7 |

Homogeneous Equations

\frac{dy}{dx} = 1 + \frac{y}{x}

homogeneous in the first-order sense?

Solution

Yes. The right-hand side can be written as a function of $y/x$:

1+\frac{y}{x} = F\!\left(\frac{y}{x}\right)

with

$$ F(v)=1+v. $$

So the equation is homogeneous.

1.8Solve a Linear Homogeneous Equation

Exam I | Problem 1.8 |

Linear First-Order Equations

Solve for $y$:

$$ y' + 3y = 0. $$

Solution

This is a linear first-order equation with $p(x)=3$ and $q(x)=0$.

The integrating factor is

\mu(x)=e^{\int 3\,dx}=e^{3x}.

Multiply through:

e^{3x}y' + 3e^{3x}y = 0.

\frac{d}{dx}(e^{3x}y)=0.

Integrate:

e^{3x}y = C,

and therefore

y = Ce^{-3x}.

1.9Identify a Bernoulli Equation

Exam I | Problem 1.9 |

Which value of $n$ makes

$$ y' + y = x y^n $$

a Bernoulli equation?

Solution

A Bernoulli equation has the form

$$ y' + p(x)y = q(x)y^n $$

with $n \neq 0,1$.

Here the equation already matches that form, with

p(x)=1, \qquad q(x)=x, \qquad n=2.

So the equation is Bernoulli with $n=2$.

1.10Interpret a Logistic Growth Model

Exam I | Problem 1.10 |

For

\frac{dP}{dt} = rP\left(1-\frac{P}{K}\right),

state the equilibrium values and describe what happens when $0<P<K$.

Solution

Set the right-hand side equal to zero:

rP\left(1-\frac{P}{K}\right)=0.

So the equilibria are

P=0 \quad \text{and} \quad P=K.

If $0<P<K$, then both factors are positive, so

\frac{dP}{dt} > 0.

That means the population increases when it is between $0$ and $K$.

Difficulty

Integrated

2.1Solve a Separable IVP

Exam II | Problem 2.1 |

Solve the initial value problem

\frac{dy}{dx} = 2xy^2, \qquad y(0)=1.

Solution

Separate variables:

\frac{1}{y^2}\,dy = 2x\,dx.

Integrate:

\int y^{-2}\,dy = \int 2x\,dx

-\frac{1}{y} = x^2 + C.

Use $y(0)=1$:

$$ -1 = C. $$

Then

-\frac{1}{y} = x^2 - 1

which gives

y = \frac{1}{1-x^2}.

2.2Use an Integrating Factor

Exam II | Problem 2.2 |

Solve for $y$:

$$ y' - 2y = e^x. $$

Solution

The integrating factor is

\mu(x)=e^{\int -2\,dx}=e^{-2x}.

Multiply through:

e^{-2x}y' - 2e^{-2x}y = e^{-x}.

\frac{d}{dx}(e^{-2x}y)=e^{-x}.

Integrate:

e^{-2x}y = -e^{-x} + C.

Multiply by $e^{2x}$:

y = -e^x + Ce^{2x}.

2.3Solve an Exact Equation

Exam II | Problem 2.3 |

Exact Equations

Solve

(3x^2y+2x)\,dx + (x^3+4y)\,dy = 0.

Solution

Let

M(x,y)=3x^2y+2x, \qquad N(x,y)=x^3+4y.

Check exactness:

M_y = 3x^2, \qquad N_x = 3x^2.

The equation is exact.

Integrate $M$ with respect to $x$:

\Psi(x,y) = \int (3x^2y+2x)\,dx = x^3y + x^2 + g(y).

Differentiate with respect to $y$:

\Psi_y = x^3 + g'(y).

Match this with $N$:

$$ x^3 + g'(y) = x^3 + 4y, $$

$$ g'(y)=4y $$

and

$$ g(y)=2y^2. $$

Thus the implicit solution is

$$ x^3y + x^2 + 2y^2 = C. $$

2.4Solve a Homogeneous Equation

Exam II | Problem 2.4 |

Solve

\frac{dy}{dx} = 1 + \frac{y}{x}.

Solution

Use the substitution

$$ y=vx, $$

\frac{dy}{dx}=v+x\frac{dv}{dx}.

Substitute into the equation:

v+x\frac{dv}{dx}=1+v.

Then

x\frac{dv}{dx}=1.

Separate and integrate:

dv = \frac{1}{x}\,dx

v=\ln|x|+C.

Since $v=y/x$,

\frac{y}{x}=\ln|x|+C

and therefore

y=x\ln|x|+Cx.

2.5Solve a Bernoulli Equation

Exam II | Problem 2.5 |

Solve

$$ y' + y = xy^2. $$

Solution

This is a Bernoulli equation with $n=2$.

Multiply by $y^{-2}$:

y^{-2}y' + y^{-1} = x.

Let

u=y^{-1}.

Then

u'=-y^{-2}y',

so the equation becomes

$$ -u' + u = x. $$

Rearrange:

$$ u' - u = -x. $$

This is linear. The integrating factor is

\mu(x)=e^{\int -1\,dx}=e^{-x}.

Multiply through:

\frac{d}{dx}(e^{-x}u)=-xe^{-x}.

Integrate:

e^{-x}u = (x+1)e^{-x} + C.

$$ u = x+1+Ce^x. $$

Since $u=1/y$,

y=\frac{1}{x+1+Ce^x}.

Also, $y=0$ is a constant solution of the original equation.

2.6Analyze an Autonomous Equation

Exam II | Problem 2.6 |

For

$$ y' = y(2-y), $$

find the equilibrium solutions and classify each as stable or unstable.

Solution

Equilibria occur when

$$ y(2-y)=0, $$

so the equilibrium values are

y=0 \quad \text{and} \quad y=2.

Check the sign of $y(2-y)$ on each interval:

y<0 \Rightarrow y(2-y)<0,

so solutions decrease below $0$.

0<y<2 \Rightarrow y(2-y)>0,

so solutions increase toward $2$.

y>2 \Rightarrow y(2-y)<0,

so solutions decrease toward $2$.

Thus $y=0$ is unstable and $y=2$ is stable.

2.7Apply the Existence-Uniqueness Theorem

Exam II | Problem 2.7 |

Existence and Uniqueness

Consider the IVP

y' = \frac{x+y}{1+y^2}, \qquad y(0)=0.

Does the existence-uniqueness theorem from the note guarantee a unique local solution?

Solution

Let

f(x,y)=\frac{x+y}{1+y^2}.

This function is continuous everywhere, and its partial derivative with respect to $y$ is also continuous because the denominator $1+y^2$ never vanishes.

So the theorem applies in a rectangle around $(0,0)$, and it guarantees a unique local solution.

2.8Interpret a Cooling Model

Exam II | Problem 2.8 |

Newton's Law of Cooling, Autonomous Equations

An object obeys

\frac{dT}{dt} = -0.2(T-18).

What is the equilibrium temperature, and what happens when $T>18$ and when $T<18$?

Solution

Set the right-hand side equal to zero:

$$ -0.2(T-18)=0. $$

So the equilibrium temperature is

$$ T=18. $$

If $T>18$, then $T-18>0$, so

\frac{dT}{dt}<0.

The object cools toward $18$.

If $T<18$, then $T-18<0$, so

\frac{dT}{dt}>0.

The object warms toward $18$.

Difficulty

Applied

3.1Build an Exponential Growth Model

Final | Problem 3.1 |

A bacterial culture has $250$ bacteria at $t=0$ and $500$ bacteria at $t=5$ hours. Assuming exponential growth, find the model $P(t)$.

Solution

Exponential growth has the form

P(t)=Ce^{kt}.

From $P(0)=250$, we get

$$ C=250. $$

Use $P(5)=500$:

500 = 250e^{5k}.

Divide by $250$:

2 = e^{5k}.

Take logs:

5k = \ln 2

k = \frac{\ln 2}{5}.

Therefore,

P(t)=250e^{(\ln 2)t/5}.

3.2Solve a Newton Cooling Problem

Final | Problem 3.2 |

Newton's Law of Cooling, Initial Value Problems

A cup of coffee starts at $90^\circ$C in a room at $20^\circ$C and satisfies

\frac{dT}{dt}=-0.3(T-20), \qquad T(0)=90.

Find $T(t)$ and the time when the coffee reaches $30^\circ$C.

Solution

The cooling model has the form

T(t)=20+Ce^{-0.3t}.

Use $T(0)=90$:

$$ 90=20+C $$

$$ C=70. $$

Thus

T(t)=20+70e^{-0.3t}.

Now set $T(t)=30$:

30=20+70e^{-0.3t}.

Subtract $20$:

10=70e^{-0.3t}.

Divide by $70$:

\frac{1}{7}=e^{-0.3t}.

Take logs:

-0.3t = \ln\!\left(\frac{1}{7}\right) = -\ln 7.

t=\frac{\ln 7}{0.3}.

3.3Write a Mixing Equation

Final | Problem 3.3 |

A tank starts with $100$ liters of brine containing $8$ grams of salt. Pure water flows in at $3$ liters per minute, and the well-mixed solution flows out at the same rate.

Let $Q(t)$ be the amount of salt in grams. Write and solve the differential equation for $Q(t)$.

Solution

Since pure water enters, the rate in is $0$.

The volume stays constant at $100$ liters, so the concentration in the tank is

\frac{Q(t)}{100}\ \text{grams per liter}.

The outflow rate of salt is

3\cdot \frac{Q}{100}=\frac{3}{100}Q.

So the differential equation is

\frac{dQ}{dt}=-\frac{3}{100}Q.

This is linear and separable. Its solution is

Q(t)=Ce^{-3t/100}.

Use $Q(0)=8$:

$$ C=8. $$

Therefore,

Q(t)=8e^{-3t/100}.

3.4Interpret a Logistic Model with Initial Data

Final | Problem 3.4 |

A population satisfies

\frac{dP}{dt}=0.4P\left(1-\frac{P}{1000}\right), \qquad P(0)=200.

Find the explicit solution.

Solution

For a logistic equation,

P(t)=\frac{K}{1+Ae^{-rt}}.

Here $K=1000$ and $r=0.4$, so

P(t)=\frac{1000}{1+Ae^{-0.4t}}.

Use $P(0)=200$:

200=\frac{1000}{1+A}.

$$ 1+A=5 $$

and

$$ A=4. $$

Therefore,

P(t)=\frac{1000}{1+4e^{-0.4t}}.

3.5Spot a Lost Constant Solution

Final | Problem 3.5 |

A student solves

\frac{dy}{dx}=xy

by dividing by $y$ and gets

y = Ce^{x^2/2}.

What solution was lost, and why does it need to be checked separately?

Solution

Dividing by $y$ assumes $y\neq 0$.

The constant solution

$$ y=0 $$

also satisfies

\frac{dy}{dx}=xy.

It must be checked separately because it is lost when the equation is divided by $y$.

Difficulty

Challenge

4.1Solve a Bernoulli Initial Value Problem

Final | Problem 4.1 |

Solve

y' + \frac{1}{x}y = xy^2, \qquad y(1)=1.

Solution

This is a Bernoulli equation with $n=2$.

Multiply by $y^{-2}$:

y^{-2}y' + \frac{1}{x}y^{-1} = x.

Let

u=y^{-1}.

Then

u'=-y^{-2}y',

-u' + \frac{1}{x}u = x.

Rearrange:

u' - \frac{1}{x}u = -x.

This is linear. The integrating factor is

\mu(x)=e^{\int -1/x\,dx}=e^{-\ln x}=\frac{1}{x}

for $x>0$.

Multiply through:

\frac{d}{dx}\!\left(\frac{u}{x}\right)=-1.

Integrate:

\frac{u}{x}=-x+C.

$$ u=-x^2+Cx. $$

Use $y(1)=1$, so $u(1)=1$:

$$ 1=-1+C $$

which gives

$$ C=2. $$

Thus

$$ u=x(2-x), $$

and since $u=1/y$,

y=\frac{1}{x(2-x)}.

4.2Solve an Exact IVP

Final | Problem 4.2 |

Solve

(2xy+1)\,dx + (x^2+2y)\,dy = 0, \qquad y(0)=1.

Solution

Let

M(x,y)=2xy+1, \qquad N(x,y)=x^2+2y.

Check exactness:

M_y=2x, \qquad N_x=2x.

So the equation is exact.

Integrate $M$ with respect to $x$:

\Psi(x,y)=\int (2xy+1)\,dx = x^2y + x + g(y).

Differentiate with respect to $y$:

\Psi_y = x^2 + g'(y).

Match with $N$:

$$ x^2 + g'(y)=x^2+2y, $$

$$ g'(y)=2y $$

and

$$ g(y)=y^2. $$

Thus

$$ x^2y + x + y^2 = C. $$

Use $y(0)=1$:

$$ 1=C. $$

So the implicit solution is

$$ x^2y + x + y^2 = 1. $$

4.3Classify Stability on a Phase Line

Final | Problem 4.3 |

For

$$ y' = y(1-y)^2, $$

find the equilibrium solutions and classify each as stable, unstable, or semistable.

Solution

Set the right-hand side equal to zero:

$$ y(1-y)^2=0. $$

So the equilibria are

y=0 \quad \text{and} \quad y=1.

Now check the sign of $y(1-y)^2$:

y<0 \Rightarrow y(1-y)^2<0,

so solutions move downward on the left of $0$.

0<y<1 \Rightarrow y(1-y)^2>0,

so solutions move upward toward $1$ from the left.

y>1 \Rightarrow y(1-y)^2>0,

so solutions move upward away from $1$ on the right.

Therefore:

y=0 \text{ is unstable,}

and

y=1 \text{ is semistable.}

4.4Compare Two Existence Questions

Final | Problem 4.4 |

For each IVP, decide whether the theorem from the note guarantees a unique local solution.

$y' = \dfrac{1}{1+y^2}, \qquad y(0)=0$
$y' = \sqrt{|y|}, \qquad y(0)=0$

Solution

For the first IVP, let

f(x,y)=\frac{1}{1+y^2}.

Both $f$ and $\partial f/\partial y$ are continuous everywhere, so the theorem guarantees a unique local solution.

For the second IVP,

f(x,y)=\sqrt{|y|}.

This function is continuous, but $\partial f/\partial y$ is not continuous at $y=0$, so the theorem does not guarantee uniqueness.

In fact, this is the kind of example in the note where non-uniqueness can occur.