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  • The number of solutions of dy/dx= y+1/x-1 when y(1) = 2 is: A. none B. one C. two D. infinite

The number of solutions of \frac{d y}{d x}=\frac{y+1}{x-1} when y(1) = 2 is:
A. none
B. one
C. two
D. infinite

Answers (1)

\begin{aligned} &\begin{array}{l} \frac{d y}{d x}=\frac{y+1}{x-1} \\ \Rightarrow \frac{d y}{y+1}=\frac{d x}{x-1} \end{array}\\ &\text { Integrate }\\ &\Rightarrow \int \frac{\mathrm{dy}}{\mathrm{y}+1}=\int \frac{\mathrm{d} \mathrm{x}}{\mathrm{x}-1} \end{aligned}

\\\log (y+1)=\log (x-1)-\log c$ \\$\log (y+1)+\log c=\log (x-1)$
using \log a+\log \mathrm{b}=\log \mathrm{ab}$
\\\log _{0} c(y+1)=\log (x-1)$ \\$\Rightarrow \frac{x-1}{y+1}=c \ldots(a)$
Now as given y(1)=2 which means when x=1, y=2 Substitute x=1 and y=2 in (a)
\\\Rightarrow \frac{1-1}{2+1}=c$ \\$C=0$ \\$\Rightarrow \frac{x-1}{y+1}=0$ \\$x-1=0$
So only one solution exists.

Option B is correct.

Posted by

infoexpert22

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