Question

The differential equation
\[\frac{dy}{dx} + Py = Q y^n , n > 2\] can be reduced to linear form by substituting

Options

• z = y^n −1

• z = yⁿ

• z = y^{n + 1}

• z = y^1 − n

Answer 1

z = y^1 − n

 

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