The number of solutions of d x d y = x − 1 y + 1 , when y ( 1 ) = 2 is

Since, d x d y = x − 1 y + 1 ⇒ y + 1 d y = x − 1 d x After integrating on both sides, we have lo g ( y + 1 ) = lo g ( x − 1 ) − lo g C C ( y + 1 ) = ( x − 1 ) C = y + 1 x − 1 If x = 1 , then y = 2 , so C = 0 Therefore, x − 1 = 0 Hence, there is only one solution.

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MathematicsDifferential EquationVariable separableEasy2 minPYQ_2023

MathematicsEasysingle choice

The number of solutions of $\frac{d y}{d x} = \frac{y + 1}{x - 1}$ , when $y (1) = 2$ is

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Answer:

Solution:

Since, $\frac{d y}{d x} = \frac{y + 1}{x - 1}$
$\Rightarrow \frac{d y}{y + 1} = \frac{d x}{x - 1}$
After integrating on both sides, we have
$lo g (y + 1) = lo g (x - 1) - lo g C$
$C (y + 1) = (x - 1)$
$C = \frac{x - 1}{y + 1}$
If $x = 1$ , then $y = 2$ , so $C = 0$
Therefore, $x - 1 = 0$
Hence, there is only one solution.

Stream:BITSATSubject:MathematicsTopic:Differential EquationSubtopic:Variable separable

⏱ 2mℹ️ Source: PYQ_2023

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