Question

Which of the following transformations reduce the differential equation \[\frac{dz}{dx} + \frac{z}{x}\log z = \frac{z}{x^2} \left( \log z \right)^2\] into the form \[\frac{du}{dx} + P\left( x \right) u = Q\left( x \right)\]

Options

• u = log x

• u = e^z

• u = (log z)⁻¹

• u = (log z)²

Answer 1

u = (log z)⁻¹

 

\[\text{Given }\frac{dz}{dx} + \frac{z}{x}\log z = \frac{z}{x^2} \left( \log z \right)^2 . . . . . . . . \left( 1 \right)\]
\[\text{ Let }u = \left( \log z \right)^{- 1} \]
\[\frac{du}{dx} = - \frac{1}{\left( \log z \right)^2} \times \frac{1}{z} \times \frac{dz}{dx}\]
\[\frac{dz}{dx} = - z \left( \log z \right)^2 \frac{du}{dx}\]
\[\text{ Substituting the value of }\frac{dz}{dx}\text{ from equation }(1)\text{ we get, }\]
\[ \therefore - z \left( \log z \right)^2 \frac{du}{dx} + \frac{z}{x}\log z = \frac{z}{x^2} \left( \log z \right)^2 \]
\[\frac{du}{dx} - \frac{1}{x}\frac{1}{\log z} = - \frac{1}{x^2}\]
\[\frac{du}{dx} - \frac{1}{x} \left( \log z \right)^{- 1} = - \frac{1}{x^2}\]
\[\frac{du}{dx} - \frac{1}{x}u = - \frac{1}{x^2}\]
It can be written as,
\[\frac{du}{dx} + p\left( x \right)u = Q\left( x \right)\]
\[\text{ where, }p\left( x \right) = - \frac{1}{x}\]
\[ q\left( x \right) = - \frac{1}{x^2}\]