If x d y d x = y ( l o g y − l o g x + 1 ) , then the solution of the equation is Options: A . &nbsp y l o g ( x y ) = c x B . &nbsp x l o g ( y x ) = c y C . &nbsp l o g ( y x ) = c x D . &nbsp l o g ( x y ) = c x

Answer: Option C : C d y d x = y x ( l o g y x + 1 ) Put y = v x ⇒ d y d x = v + x d v d x ∴ v + x d v d x = v l o g v + v ⇒ 1 v l o g v d v = 1 x d x ⇒ ∫ 1 v l o g v d v = ∫ 1 x d x ⇒ l o g ( l o g v ) = l o g x + l o g c ⇒ l o g y x = c x

If xdydx=y(log y−log x+1), then the solution of the equation

Discussion Forum : Differential Equations

Question -

If $x \frac{d y}{d x} = y (l o g y - l o g x + 1)$ , then the solution of the equation is

Options:

A .

y l o g (\frac{x}{y}) = c x

B .

x l o g (\frac{y}{x}) = c y

C .

l o g (\frac{y}{x}) = c x

D .

l o g (\frac{x}{y}) = c x

Answer: Option C
:
C

\frac{d y}{d x} = \frac{y}{x} (l o g \frac{y}{x} + 1)

Put

y = v x \Rightarrow \frac{d y}{d x} = v + x \frac{d v}{d x}

∴ v + x \frac{d v}{d x} = v l o g v + v \Rightarrow \frac{1}{v l o g v} d v = \frac{1}{x} d x \Rightarrow \int \frac{1}{v l o g v} d v = \int \frac{1}{x} d x \Rightarrow l o g (l o g v) = l o g x + l o g c

\Rightarrow l o g \frac{y}{x} = c x

Was this answer helpful ?

Next Question

Discussion Forum : Differential Equations

Submit Your Solution hear: