First Order ODEs Practice

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

Rewrite it in standard linear form:

This matches

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

Substitute the point into the right-hand side:

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

Integrate both sides with respect to $x$:

So the general solution is

Integrate the differential equation:

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

so

Therefore,

Equilibria occur when the right-hand side is zero:

So

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

Let

Compute the partial derivatives:

Since these are equal, the equation is exact.

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

with

So the equation is homogeneous.

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

The integrating factor is

Multiply through:

So

Integrate:

and therefore

A Bernoulli equation has the form

with $n \neq 0,1$.

Here the equation already matches that form, with

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

Set the right-hand side equal to zero:

So the equilibria are

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

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

Separate variables:

Integrate:

so

Use $y(0)=1$:

Then

which gives

The integrating factor is

Multiply through:

So

Integrate:

Multiply by $e^{2x}$:

Let

Check exactness:

The equation is exact.

Integrate $M$ with respect to $x$:

Differentiate with respect to $y$:

Match this with $N$:

so

and

Thus the implicit solution is

Use the substitution

so

Substitute into the equation:

Then

Separate and integrate:

so

Since $v=y/x$,

and therefore

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

Multiply by $y^{-2}$:

Let

Then

so the equation becomes

Rearrange:

This is linear. The integrating factor is

Multiply through:

Integrate:

So

Since $u=1/y$,

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

Equilibria occur when

so the equilibrium values are

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

so solutions decrease below $0$.

so solutions increase toward $2$.

so solutions decrease toward $2$.

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

Let

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.

Set the right-hand side equal to zero:

So the equilibrium temperature is

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

The object cools toward $18$.

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

The object warms toward $18$.

Exponential growth has the form

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

Use $P(5)=500$:

Divide by $250$:

Take logs:

so

Therefore,

The cooling model has the form

Use $T(0)=90$:

so

Thus

Now set $T(t)=30$:

Subtract $20$:

Divide by $70$:

Take logs:

So

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

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

The outflow rate of salt is

So the differential equation is

This is linear and separable. Its solution is

Use $Q(0)=8$:

Therefore,

For a logistic equation,

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

Use $P(0)=200$:

So

and

Therefore,

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

The constant solution

also satisfies

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

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

Multiply by $y^{-2}$:

Let

Then

so

Rearrange:

This is linear. The integrating factor is

for $x>0$.

Multiply through:

Integrate:

So

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

which gives

Thus

and since $u=1/y$,

Let

Check exactness:

So the equation is exact.

Integrate $M$ with respect to $x$:

Differentiate with respect to $y$:

Match with $N$:

so

and

Thus

Use $y(0)=1$:

So the implicit solution is

Set the right-hand side equal to zero:

So the equilibria are

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

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

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

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

Therefore:

and

For the first IVP, let

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

For the second IVP,

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.