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Q45:
The number of solutions of $\frac{dy}{dx} = \frac{y+1}{x-1}$ when $y (1) = 2$ is :

Solution :

Step 1: Identifying the Differential Equation Type

The given equation is $\frac{dy}{dx} = \frac{y+1}{x-1}$. This is a first-order linear ordinary differential equation, which is also separable. We can rearrange the variables as follows: $\frac{1}{y+1} dy = \frac{1}{x-1} dx$ [Per the Method of Separation of Variables].

Step 2: Integrating Both Sides

Integrate both sides with respect to their variables: $\int \frac{1}{y+1} dy = \int \frac{1}{x-1} dx$ Applying the standard integration rule $\int \frac{1}{u} du = \ln|u| + C$, we obtain: $\ln|y+1| = \ln|x-1| + C_1$ [By the Fundamental Theorem of Calculus].

Step 3: Solving for the General Solution

To isolate $y$, exponentiate both sides: $|y+1| = e^{\ln|x-1| + C_1}$ $|y+1| = e^{C_1} \cdot |x-1|$ Let $e^{C_1} = k$ (where $k > 0$). Removing the absolute value signs introduces a constant $C$ that can take any real value: $y + 1 = C(x - 1)$ $y = C(x - 1) - 1$

Step 4: Applying the Initial Condition

We are given the initial condition $y(1) = 2$. Substituting $x = 1$ and $y = 2$ into the general solution: $2 = C(1 - 1) - 1$ $2 = C(0) - 1$ $2 = -1$

Step 5: Analyzing the Consistency

The resulting equation $2 = -1$ is a logical contradiction. This indicates that there is no value of the constant $C$ that satisfies the condition $y(1) = 2$ given the differential equation structure. Geometrically, the function $y = C(x - 1) - 1$ describes a family of lines passing through the point $(1, -1)$. The condition $y(1) = 2$ requires the solution curve to pass through $(1, 2)$. Since the point $(1, 2)$ does not lie on any line in the family $y = C(x - 1) - 1$, no solution exists.

(1, 2) (1, -1)

Final Solution: The number of solutions is 0.


More Questions from Class 12 Mathematics Differential Equations EXERCISE 9.3


CBSE Solutions for Class 12 Mathematics Differential Equations


Chapters in CBSE - Class 12 Mathematics


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