Test Index

AP EAMCET Engineering 18 Sep 2020 Shift 2 Paper

Section: Mathematics

© examsnet.com

Question : 75 of 160

Marks: +1, -0

The solution of

x\frac{dy}{dx}=y(\log y-\log x+1)

is

$y=xe^{cx}$
$y^2=cx^2$
$y^2=cx\log(x)$
$\log(y)=cx$

Solution:

x \frac{dx}{dx}=y(\log y-\log x+1)

\frac{dy}{dx}=\frac{y}{x}\left(\log \frac{y}{x}+1\right)

...(i)

Let

f(x, y)=\frac{y}{x}\left(\log \frac{y}{x}+1\right)

f(kx, ky)=\frac{ky}{kx}\left(\log \frac{ky}{kx}+1\right)

f(kx, ky)=\frac{y}{x}\left(\log \frac{y}{x}+1\right)=f(x, y)

\therefore

Given differentiate Equation is homogeneous

\left\{\begin{array}{l}\therefore \text{ put, } y=v x \\ \frac{dy}{dx}=v+x\frac{dv}{dx}\end{array}\right.

From Eq. (i),

v+x\frac{dv}{dx}=\frac{vx}{x}\left(\log \frac{vx}{x}+1\right)

v+x\frac{dv}{dx}=v(\log v+1)

v+x\frac{dv}{dx}=v\log v+v

\frac{1}{v\log v} dv = \frac{1}{x} dx

Integrating on both sides,

\int \frac{1}{v} \cdot \frac{1}{\log v} dv = \int \frac{1}{x} dx

Put,

\log v = t

\frac{1}{v} dv = dt

\int \frac{1}{t} dt = \int \frac{1}{x} dx

\log t = \log x + \log c

\log(\log v) = \log x c

\log_{e} v = x c

v=e^{xc}

\frac{y}{x}=e^{xc}

y=x \cdot e^{xc}

© examsnet.com

Go to Question:

Prev Question

Next Question