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Homogeneous ODE

Homogeneous ODE solver. Enter F(v) and an initial value to substitute v=y/x, separate variables and solve numerically—with steps.

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dy/dx = F(y/x), set v = y/x
RHS in terms of v
initial x (>0)
initial y(x₀)
target point

📖 Tutorial｜Homogeneous ODE

1. Definition
If dy/dx can be written as a function of y/x alone, F(y/x), the equation is homogeneous. Substituting v=y/x makes it separable.
2. Symbols
SymbolMeaning
vnew variable y/x
F(v)RHS as a function of v
y = vxsubstitution relation
3. How this tool works
  • Set v = y/x, so y = vx and dy/dx = v + x·dv/dx;
  • Then v + x dv/dx = F(v), i.e. dv/(F(v)−v) = dx/x;
  • This is separable; integrate and substitute back v=y/x;
  • This tool applies Euler directly to dy/dx=F(y/x).
4. Steps
  1. Rewrite the RHS as F(y/x);
  2. Enter F(v), initial values and target x₁;
  3. Click Calculate to see the substitution and numerical answer.
5. Example
Example: dy/dx = y/x, y(1)=2. With v=y/x: v+x dv/dx = v, so dv/dx=0, v=C=2, y=2x. Numerical y(2)=4 matches exactly.
6. Pitfalls
The RHS must be expressible as a function of y/x alone;
x₀ must not be 0 (y/x is undefined);
Remember to substitute back v=y/x, not treat v as y.

❓ FAQ｜Homogeneous ODE

When is an equation homogeneous?
When the RHS is a function of y/x alone, i.e. all terms have the same degree.
Why substitute v=y/x?
After substitution the equation contains only v and x and becomes separable.
Why can't x₀ be 0?
Because y/x is undefined at x=0 and the substitution breaks down.
Is the answer exact?
This tool uses Euler for a numerical approximation; integrate by hand for an exact answer.
Which functions are supported?
F(v) is written with v; common elementary functions; never omit the multiplication sign.