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

\[\left(3y^{2}-xy \right)\, dx + x^{2}\,dy = 0\]

using algebra to isolate the differential

\[\left(3y^{2}-xy\right)\,dx = -x^{2}\,dy\]

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

Splitting the fraction and canceling terms

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

using the \(v\)-substitution

\[v = \frac{y}{x} \qquad y = vx\]

\[\frac{dy}{dx} = v+x \frac{dv}{dx}\]

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

\[\frac{dx}{dt} = \frac{x}{t} + \frac{\sqrt{t^{2}\left (8+ \frac{x^{2}}{t^{2}}\right)}}{2x}\]

This lets us pull out the \(t^{2}\) from the radical giving us

\[\frac{dx}{dt} = \frac{x}{t} + \frac{t\sqrt{\left (8+ \frac{x^{2}}{t^{2}}\right)}}{2x}\]

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

\[\frac{dx}{dt} = \frac{x}{t} + \frac{\sqrt{\left (8+ \frac{x^{2}}{t^{2}}\right)}}{\frac{2x}{t}}\]

Now we can use the substitutions

\[v = \frac{x}{t} \qquad tv = x\]

\[\frac{dx}{dt} = v + t \frac{dv}{dt}\]

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

\[u = 8+v^{2}\qquad du = 2v\,dv\]

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

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

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

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