By E. Z.
\(\ y'+p(x)y=q(x)y^n \)
\(\ y'y^{-n}+p(x)y^{1-n}=q(x) \)
\(\ z=y^{1-n}, z'=(1-n)y^{-n}y' \)
\(\ z'+p(x)(1-n)z=q(x)(1-n) \)
For a known particular solution \(\ y_1 \)
\(\ y' = r(x) + p(x) y + q(x) y^2 \)
\(\ y(x)=y_1(x)+ \frac{1}{z(x)}, y'(x)=y_1'(x)-\frac {z'(x)}{z^2(x)} \)
\(\ z'+z[p(x)+2q(x)y_1]=-q(x) \)
and then transforming it back into 𝓎 from 𝓏\(\ \frac{∂ φ(x,y)}{∂x} = M(x,y) \)
\(\ ∫M(x,y)dx+g(y)=ψ(x,y) \)
\(\ \frac{∂ ψ}{∂ y}=N(x,y) \)
So that linear equations are easy to solve unlike non-linear equations we can approximate non-linear terms to linear terms for some values of variable.
In order to find spesific \(\ c_1 \) & \(\ c_1 \) there is needed for 2 additional conditions.