Question

The differential equation $\frac{dy}{dx}+Py=Q{y}^n,n>2$ can be reduced to linear form by substituting
(a) z = y^{n −1}
(b) z = yⁿ
(c) z = y^{n + 1}
(d) z = y^{1 − n}

Answer 1

(d) z = y^{1 − n}

We have,
$$\begin{gathered} \frac{dy}{dx}+Py=Q{y}^n \\ \Rightarrow y^{-n}\frac{dy}{dx}+P{y}^{1-n}=Q.....\left(1\right) \\ \mathrm{Put}z=y^{1-n} \\ \mathrm{Integrating}\mathrm{both}\mathrm{sides}\mathrm{with}\mathrm{respect}\mathrm{to}x,\mathrm{we}\mathrm{get} \\ \frac{dz}{dx}=\left(1-n\right)y^{-n}\frac{dy}{dx} \\ \Rightarrow y^{-n}\frac{dy}{dx}=\frac{1}{\left(1-n\right)}\frac{dz}{dx} \\ \mathrm{Now},\left(1\right)\mathrm{becomes} \\ \frac{1}{\left(1-n\right)}\frac{dz}{dx}+Pz=Q \\ \Rightarrow \frac{dz}{dx}+P\left(1-n\right)z=Q\left(1-n\right) \\ \mathrm{Which}\mathrm{is}\mathrm{linear}\mathrm{form}\mathrm{of}\mathrm{differential}\mathrm{equation}. \\ \mathrm{Therefore},\mathrm{the}\mathrm{given}\mathrm{differential}\mathrm{equation}\mathrm{can}\mathrm{be}\mathrm{reduce}\mathrm{to}\mathrm{linear}\mathrm{form}\mathrm{by}\mathrm{the}\mathrm{substitution}, \\ z=y^{1-n} \end{gathered}$$