Question

The differential equation of the family of straight lines whose slope is equal to y-intercept, is

• A. $(\mathrm{x}+1)\frac{d\mathrm{y}}{d\mathrm{x}}–\mathrm{y}=0$
• B. $(\mathrm{x}+1)\frac{d\mathrm{y}}{d\mathrm{x}}+\mathrm{y}=0$
• C. $\frac{d\mathrm{y}}{d\mathrm{x}}=\frac{(\mathrm{x}-1)}{(\mathrm{y}-1)}$
• D. $\frac{d\mathrm{y}}{d\mathrm{x}}=\frac{(\mathrm{x}+1)}{(\mathrm{y}+1)}$

Answer 1

The correct option is A

$(\mathrm{x}+1)\frac{d\mathrm{y}}{d\mathrm{x}}–\mathrm{y}=0$

Explanation for the correct option:

Given, slope is equal to y-intercept

General equation of straight line is: $\mathrm{y}=\mathrm{mx}+\mathrm{c}$

Given $\Rightarrow \mathrm{m}=\mathrm{c}$ ( as slope is equal to y-intercept)

Therefore, the equation becomes

$\mathrm{y}=\mathrm{cx}+\mathrm{c}\to \left(1\right)$

On differentiating with respect to $x$,

$$\begin{gathered} \frac{d\mathrm{y}}{d\mathrm{x}}=\mathrm{c} \\ \Rightarrow {\mathrm{y}}_1=\mathrm{c}\left(\because \frac{d\mathrm{y}}{d\mathrm{x}}={\mathrm{y}}_1\right) \end{gathered}$$

On substituting ${\mathrm{y}}_1=\mathrm{c}=\mathrm{m}$ in equation $\left(1\right)$ we get,

$\begin{array}{rcl}\mathrm{y}& =& {\mathrm{y}}_1\mathrm{x}+{\mathrm{y}}_1\\ & \Rightarrow & \left(\mathrm{x}+1\right){\mathrm{y}}_1-\mathrm{y}=0\\ & \Rightarrow & \left(\mathrm{x}+1\right)\frac{d\mathrm{y}}{d\mathrm{x}}-\mathrm{y}=0\end{array}$

Hence the correct answer is option (A)