Differentiation, Differential Equations, Ode, Mathematics, Derivatives.
u = y¹⁻ⁿ By using chain rule we get du/dx = (1 - n)y⁻ⁿ. dy/dx which can also be written as y ⁻ⁿ. dy/dx = du/dx . 1/(1 - n) After substitution we get du/dx . 1/(1 - n) + P(x)u = Q(x) By algebraic manipulation we get du/dx + (1 - n)P(x)u = (1 - n)Q(x) which can be solved as a linear differential