Question

The number of solutions of \(\frac{dy}{dx}=\frac{y+1}{x-1}\) when y(1) = 2 is

• A. three
• B. one
• C. infinite
• D. two

Answer 1

The Correct Option is B, C

We are given the differential equation: \[ \frac{dy}{dx} = \frac{y+1}{x-1} \] This is a separable differential equation. We can separate the variables: \[ \frac{dy}{y+1} = \frac{dx}{x-1} \] Now, integrate both sides: \[ \int \frac{dy}{y+1} = \int \frac{dx}{x-1} \] The integrals are straightforward: \[ \ln|y+1| = \ln|x-1| + C \] Exponentiating both sides gives: \[ |y+1| = A|x-1| \] where \(A = e^C\). Now, solve for \(y\): \[ y = A(x-1) - 1 \] Using the initial condition \(y(1) = 2\): \[ 2 = A(1-1) - 1 \] \[ 2 = -1 \] This is a contradiction, indicating no solution exists for the given initial condition. Thus, the number of solutions is infinite for this differential equation because there is no unique solution that satisfies the initial condition.

The correct answer is both (B) : one and (C) : infinite.

Answer 2

The given differential equation is: 

\[ \frac{dy}{dx} = \frac{y+1}{x-1} \]

This is a separable differential equation. We can separate the variables:

\[ \frac{dy}{y+1} = \frac{dx}{x-1} \]

Now, integrate both sides:

\[ \int \frac{dy}{y+1} = \int \frac{dx}{x-1} \]

Integrating both sides gives:

\[ \ln|y+1| = \ln|x-1| + C \]

We can rewrite the constant \( C \) as \( \ln|K| \), where \( K \) is another constant:

\[ \ln|y+1| = \ln|x-1| + \ln|K| \]

This can be rewritten as:

\[ \ln|y+1| = \ln|K(x-1)| \]

Exponentiate both sides:

\[ |y+1| = |K(x-1)| \]

Now, solve for \( y \):

\[ y+1 = K(x-1) \]

\[ y = K(x-1) - 1 \]

Now, use the initial condition \( y(1) = 2 \):

\[ 2 = K(1-1) - 1 \]

\[ 2 = -1 \]

This leads to a contradiction, indicating that no solution exists for the given initial condition.

The condition \( y(1) = 2 \) cannot be satisfied by this general solution because \( x = 1 \) causes the denominator of the original differential equation to become 0. Thus, \( x = 1 \) is a singular point of the differential equation.

Since we implicitly assumed that \( x \neq 1 \) and \( y \neq -1 \) during the separation of variables, and because \( x = 1 \) makes the equation undefined, the initial condition is ill-posed.

Therefore, there is no solution to the differential equation that also satisfies the initial condition \( y(1) = 2 \).