Question

The solution curve of the differential equation x + y(dy/dx) = 2y makes intercept on y - axis of length

• A. 0
• B. c
• C. $\infty$
• D. None of these

Answer 1

Answer: (B)

c

Answer 2

$x + y\frac{dy}{dx} = 2y \Rightarrow \frac{dy}{dx} = 2y - x \Rightarrow \frac{dy}{dx} = \frac{2y - x}{y}$

Let $y = \nu x \Rightarrow \frac{dy}{dx} = \nu + x\frac{d\nu}{dx}$

$\therefore$ given equation becomes $\nu + x\frac{d\nu}{dx} = \frac{2\nu - 1}{\nu}$

$\Rightarrow$ $x\frac{d\nu}{dx} = \frac{2\nu - 1}{\nu} - \nu = \frac{2\nu - 1 - \nu^{2}}{\nu} = \frac{- (\nu - 1)^{2}}{\nu}$

$\Rightarrow - \frac{\nu}{(\nu - 1)^{2}}d\nu = \frac{dx}{x} \Rightarrow 0 = \frac{dx}{x} + \frac{\nu}{(\nu - 1)^{2}}d\nu$

Integrate both sides, we get $k = \log x + \int_{}^{}\frac{\nu}{(\nu - 1)^{2}}d\nu$

$\Rightarrow k = \log x + \int_{}^{}\frac{t + 1}{t^{2}}dt$ $(putt = \nu - 1 \Rightarrow = d\nu)$

$\Rightarrow k = \log x + \int_{}^{}{\left( \frac{1}{t} + \frac{1}{t^{2}} \right)dt} \Rightarrow k = \log x + \log t - \frac{1}{t}$

$\Rightarrow k = \log x + \log\left( \frac{y}{x} - 1 \right) - \frac{1}{\frac{y}{x} - 1}$

$\Rightarrow k = \log x + \log(y - x) - \log x - \frac{x}{y - x}$

$\Rightarrow$ $\log(y - x) - \frac{x}{y - x} = k$

At x = 0, k = log_e y $\Rightarrow$ y = e^k = c.