For dy dx=f(yx), use y=vx, where v is a function of x and x 0. dy dx=v+x dv dx, x dv dx=f(v)-v. Separate the variables. Check constant values of v satisfying f(v)=v before dividing; these give straight-line solutions. Substitute v= yx back into the result. Example For dy dx=1+ yx, substitution gives x dv dx=1. v=ln|x|+C, y=x(ln|x|+C), on an interval not crossing zero.

For \(\dfrac{\mathrm dy}{\mathrm dx}=\mathrm{f}\left(\dfrac yx\right)\), use \(y=vx\), where \(v\) is a function of \(x\) and \(x\ne0\).

\(\dfrac{\mathrm dy}{\mathrm dx}=v+x\dfrac{\mathrm dv}{\mathrm dx},\qquad x\dfrac{\mathrm dv}{\mathrm dx}=\mathrm{f}(v)-v\).

Separate the variables. Check constant values of \(v\) satisfying \(\mathrm{f}(v)=v\) before dividing; these give straight-line solutions. Substitute \(v=\dfrac yx\) back into the result.

For \(\dfrac{\mathrm dy}{\mathrm dx}=1+\dfrac yx\), substitution gives \(x\dfrac{\mathrm dv}{\mathrm dx}=1\).

\(v=\ln|x|+C,\qquad y=x(\ln|x|+C)\), on an interval not crossing zero.

Differential Equations

International Baccalaureate Mathematics: Analysis and Approaches Higher Level

Homogeneous Differential Equations

Homogeneous Differential Equations