Lesson 8 - Substitutions and Transformations for First Order Equations

We will look to three different forms of equations. This will cover a lot of ground. I recommend looking at each part A, B, and C individually and becoming confident in one before moving to the next.

A) Homogeneous Equations
B) Equations of the form $\frac{dy}{dx}=f(ax+by)$
C) Bernoulli Equations

A)

Homogeneous equations can be written in the form: $\frac{dy}{dx}=f\left(\frac{y}{x}\right)$

We can use the substitution $y=vx$ to solve the DE.

Differentiate $y=vx$ with respect to x.

We can substitute these into the equation, which will allow us to separate variables.

Example

We first need to convert this into a form that we can work with. We can do this by multiplying the numerator and the denominator by $\frac{1}{x}$.

Now we can solve using our substitutions from before.

Starting with the right side:

Left side:

Now, we can separate variables to solve this DE using a method that we are already comfortable with.

The left side of the equation can be simply solved using the known integral $\int \frac{1}{x}$ = $\ln |x|$. The right hand side can be solved using the u-substitution method of integration. I will skip this step.

We end up with

At this point, we can multiple the entire equation by $2$ to get rid of the fraction. Notice that $2C$ can be rewritten as $C_1$ because the $2C$ is still a constant, and the $2$ is absorbed by the constant.

Let's free up the variables.

$-\ln |a|=\ln |\frac{1}{a}|$
$e^{C_1}=C_2$ for the same reason as before.

We can now sub back in $v=\frac{y}{x}$.

Try for yourself:

1. $x^2{y}^{\prime }-(xy+y^2+x^2)=0$
2. $(x^2+y^2)dx-2xydy=0$

B)

Equations of the form $\frac{dy}{dx}=f(ax+by)$

We use the substitution $z=ax+by$.

Differentiate $z$ with respect to $x$.

Example

Substitute $z=(x-y)$

Since $a=1$ and $b=-1$, $\frac{dz}{dx}=1-\frac{dy}{dx}$.

Now, we substitute:

This now becomes a separable differential equation.

We can expand out the right integral:

We can now factor the denominator:

Using partial fractions, we can split the fraction to integrate easier:

Using $z=-4$:

Using $z=-6$:

Pulling the constants out front, we have:

Using log property $\ln |a|-\ln |b|=\ln |\frac{a}{b}|$, we can condense this down.

The left side of the equation is much more trivial:

Now:

C)

A first order differential equation is said to be a Bernoulli equation if it can be written in the form:

What do we notice about this form? It's conveniently similar to a linear first order equation. The goal is to achieve that form by ridding ourselves of the $y^n$ that is attached to Q(x).

To do so, we use the substitution $v=y^{1-n}$.

Taking the derivative of $v$:

One more manipulation gives us:

Steps to solving a Bernoulli equation:

1. Determine $n$.
2. Set up substitutions.
3. Multiply entire equation on both sides by $y^{-n}$.
4. Convert to linear first order equation.
5. Solve using integrating factor.

Example

To solve, let's determine our value of $n$. Since $Q(x)y^n=2x^2{y}^4$, $n=4$.

Let's set up our substitutions:

Now let's multiply the entire equation by $y^{-n}$.

Now, we can utilize our substitutions that we created earlier.

Let's now manipulate this into standard form by multiplying by $-3$ across.

You can now continue solving this equation like you would in Lesson 6 - Linear First-Order Equations. I will skip this step.

We arrive at:

We substitute back $y$ into $v$.

Solving for the initial condition $y(1)=3$, you will get:

Try for yourself:

1. $y^{-3}-5y^{-2}=-\frac{5}{2}x$
2. $3x{y}^2{y}^{\prime }=3x^4+y^3$

Next Lesson: Lesson 9 - Homogeneous Second Order Linear Equations

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