Q76 / 180 · COMEDK 2025 (10 May Shift 3) · Mathematics
The number of solutions of $\dfrac{d y}{d x}=\dfrac{y+1}{x-1}$, when $y(1)=2$ is
- Aone
- Binfinite
- Ctwo
- Correct answerDzero
Solution
The given differential equation is a first-order linear separable differential equation: $\dfrac{dy}{dx} = \dfrac{y+1}{x-1}$.
Separating the variables, we have $\int \dfrac{dy}{y+1} = \int \dfrac{dx}{x-1}$.
Integrating both sides, we get $\ln|y+1| = \ln|x-1| + C$, which can be written as $y+1 = k(x-1)$ where $k$ is a constant.
The initial condition is given as $y(1) = 2$. Substituting $x=1$ and $y=2$ into the equation $y+1 = k(x-1)$, we get $2+1 = k(1-1)$, which simplifies to $3 = k(0)$.
Since $3 = 0$ is a contradiction, there is no value of $k$ that satisfies the initial condition $y(1) = 2$.
Therefore, there are zero solutions to the given differential equation with the specified initial condition.
Answer: zero
Separating the variables, we have $\int \dfrac{dy}{y+1} = \int \dfrac{dx}{x-1}$.
Integrating both sides, we get $\ln|y+1| = \ln|x-1| + C$, which can be written as $y+1 = k(x-1)$ where $k$ is a constant.
The initial condition is given as $y(1) = 2$. Substituting $x=1$ and $y=2$ into the equation $y+1 = k(x-1)$, we get $2+1 = k(1-1)$, which simplifies to $3 = k(0)$.
Since $3 = 0$ is a contradiction, there is no value of $k$ that satisfies the initial condition $y(1) = 2$.
Therefore, there are zero solutions to the given differential equation with the specified initial condition.
Answer: zero
Hence, the correct answer is option: 'D'.