First order homogeneous ordinary differential equations

Next type of equation were going to solve is a Homogeneous Equation

These equations have the form

$$\frac{dy}{dx}=F\left(\frac{y}{x}\right)$$

The Two Requirements for it to be a homogeneous equation are
- All degrees of the variables are the same
- all variables are in the form of$\frac{y}{x}$

Once we get it into its "standard" form (rearranged to meet the above criteria) we're going to use the substitution

$$v=\frac{y}{x}$$

and rearranged for $y$ we get

$$y=vx$$

then using implicit differentiation and the product rule

$$\frac{d}{dx}(y=vx)$$

$$\Rightarrow$$

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

substitute the $\frac{dy}{dx}$ into the original equation

$$\frac{dy}{dx}=F\left(\frac{y}{x}\right)$$

$$\Rightarrow$$

$$v+x\frac{dv}{dx}=F(v)$$

with some algebra we can rearrange this into a first order separable ODE

$$v+x\frac{dv}{dx}=F(v)$$

$$F(v)-v=x\frac{dv}{dx}$$

$$\frac{F(v)-v}{x}=\frac{dv}{dx}$$

$$\frac{dx}{x}=\frac{dv}{F(v)-v}$$

$$\int \frac{dx}{x}=\int \frac{dv}{F(v)-v}$$

Don't forget to re-substitute your $\frac{y}{x}$'s back in for your $v$'s after integration

---

Examples

$$(3y^2-xy)dx+x^{2}dy=0$$

using algebra to isolate the differential

$$(3y^2-xy)dx=-x^{2}dy$$

$$\frac{dy}{dx}=\frac{-3y^2-xy}{x^2}$$

Splitting the fraction and canceling terms

$$\frac{dy}{dx}=\frac{-3y^2}{x^2}-\frac{y}{x}$$

using the $v$-substitution

$$v=\frac{y}{x}y=vx$$

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

and substituting it in

$$v+x\frac{dv}{dx}=-3v^2+v$$

subtracting the $v$ from a side cancels it giving us

$$x\frac{dv}{dx}=-3v^2$$

this can be a separable equation, rearanging it with algebra and integrating...

$$\int -\frac{dv}{3v^2}=\int \frac{dx}{x}$$

$$-\frac{1}{3v}=-\ln |x|+c$$

substituting $\frac{y}{x}$ back in for $v$

$$-\frac{x}{3y}=-\ln |x|+c$$

Finally rearranging it into an implicit form by solving for $y$, we get our final answer

$$y=\frac{x}{3\ln |x|+c}$$

$$\frac{dx}{dt}=\frac{2x^2+t\sqrt{8t^2+x^2}}{2tx}$$

Here we're going to fully go through an example that is difficult to get into the correct form, lets start by separating the fraction

$$\frac{dx}{dt}=\frac{2x^2}{2tx}+\frac{t\sqrt{8t^2+x^2}}{2tx}$$

canceling terms gives us

$$\frac{dx}{dt}=\frac{x}{t}+\frac{\sqrt{8t^2+x^2}}{2x}$$

next to deal with the squared terms in the radical were going to factor out a $t^2$ from both terms. When we are factoring we are just dividing by the term we are 'pulling out' so we get

$$\frac{dx}{dt}=\frac{x}{t}+\frac{\sqrt{t^2(8+\frac{x^2}{t^2})}}{2x}$$

This lets us pull out the $t^2$ from the radical giving us

$$\frac{dx}{dt}=\frac{x}{t}+\frac{t\sqrt{(8+\frac{x^2}{t^2})}}{2x}$$

and finally using the property $a/(b/c)=(ac/b)$ in the reverse order¹ we would normally do, we can bring the $t$ into the denominator of the denominator giving us

$$\frac{dx}{dt}=\frac{x}{t}+\frac{\sqrt{(8+\frac{x^2}{t^2})}}{\frac{2x}{t}}$$

Now we can use the substitutions

$$v=\frac{x}{t}tv=x$$

$$\frac{dx}{dt}=v+t\frac{dv}{dt}$$

Substituting this in we get

$$v+t\frac{dv}{dt}=v+\frac{\sqrt{8+v^2}}{2v}$$

subtracting a $v$ gives us

$$t\frac{dv}{dt}=\frac{\sqrt{8+v^2}}{2v}$$

We now have a separable equation, using algebra to split the differential and integrating

$$\int \frac{2v}{\sqrt{8+v^2}}dv=\int \frac{dt}{t}$$

to solve the left side we'll use the $u$-substitution method

$$u=8+v^{2}du=2vdv$$

$$\int \frac{du}{u^{\frac{1}{2}}}=\ln |t|+c$$

$$2\sqrt{u}=\ln |t|+c$$

re-subbing back in the $v$ from the $u$-sub

$$2\sqrt{8+v^2}=\ln |t|+c$$

re-subbing back in the $\frac{x}{t}$ from $v$-sub

$$2\sqrt{8+{\left(\frac{x}{t}\right)}^2}=\ln |t|+c$$

Now, finally we have our answer given in explicit form

---

<-- Bernoulli ODE

Differential Equation Index
Home