Homogeneous substitution
\[y=vx,\quad \frac{dy}{dx}=v+x\frac{dv}{dx}\]
Differential Equations
Homogeneous Differential Equations
Understand
A function \(F(x,y)\) is homogeneous of degree \(n\) if \(F(\lambda x,\lambda y)=\lambda^nF(x,y)\). The equation \(\frac{dy}{dx}=F(x,y)\) is homogeneous when \(F\) has degree 0, i.e. \(F\) depends only on \(\frac yx\).
Method: put \(y=vx\), so \(\frac{dy}{dx}=v+x\frac{dv}{dx}\). The equation becomes separable in \(v\) and \(x\). Integrate, then replace \(v\) by \(\frac yx\).
Example: \(\frac{dy}{dx}=\frac{x+y}{x}\). With \(y=vx\): \(v+x\frac{dv}{dx}=1+v\Rightarrow dv=\frac{dx}{x}\Rightarrow v=\log|x|+C\Rightarrow y=x\log|x|+Cx\).
If the equation is easier in the form \(\frac{dx}{dy}=G\left(\frac xy\right)\), put \(x=vy\) instead.
Key Concepts
- Test: replace x, y by λx, λy; λ must cancel.
- Substitute y = vx, dy/dx = v + x dv/dx.
- Separate v and x, integrate, back-substitute v = y/x.
Formula Bank
Solved Examples
Homogeneous equation
Solve \(\frac{dy}{dx}=\frac{y}{x}+\frac{x}{y}\).
\(y=vx\): \(x\frac{dv}{dx}=\frac1v\Rightarrow v\,dv=\frac{dx}x\Rightarrow\frac{v^2}2=\log|x|+C\Rightarrow y^2=2x^2(\log|x|+C)\).
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Check homogeneity, substitute y = vx, separate, integrate, replace v.
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