First slide

Methods of solving first order first degree differential equations

Question

$Solution of differential equation \frac{d z}{d x} + \frac{z}{x} \log z = \frac{z}{x^{2}} (\log z)^{2} is$

Moderate

A

$\frac{x}{\log z} = \frac{1}{2} + c x^{2}$
B

$\frac{x}{\log z} = \frac{1}{2} - c x^{2}$
C

$\frac{x}{\log z} = \frac{1}{3} + c x^{2}$
D

none of these

Solution

Dividing the complete equation by $z {(\log z)}^{2}$ the transformed equation obtained is
$\begin{array}{l} \frac{1}{z (\log z)^{2}} \frac{d z}{d x} + \frac{1}{x \log z} = \frac{1}{x^{2}}; Now substituting \\ \frac{1}{\log z} = y \Rightarrow - \frac{1}{2 (\log z)^{2}} \frac{d z}{d x} = \frac{d y}{d x} \dots \dots \dots . . (i) \\ Then (i) reduced to linear differential equation as below \\ - \frac{d y}{d x} + \frac{1}{x} y = \frac{1}{x^{2}} \\ \Rightarrow \frac{d y}{d x} - \frac{1}{x} y = - \frac{1}{x^{2}} and I . F = e^{- 1 \int \frac{1}{x} d x} = \frac{1}{x} \end{array}$
$\begin{array}{l} Its solution is given by \frac{y}{x} = - \int \frac{1}{x^{3}} d x + c \\ \Rightarrow \frac{y}{x} = \frac{1}{2 x^{2}} + c \Rightarrow \frac{y}{x} = \frac{1}{2 x^{2}} + c \\ \Rightarrow \frac{x}{\log z} = \frac{1}{2} + c x^{2} is the required general solution \end{array}$

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