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Q49:
The differential equation $y\frac{dy}{dx} + x = c$ represents :

Solution :

Step 1: Identifying the Differential Equation

The given differential equation is:

$y \frac{dy}{dx} + x = c$

where $c$ is an arbitrary constant. This is a first-order, first-degree differential equation. [By definition of Separable Variables differential equations].

Step 2: Separation of Variables

Rearrange the equation to isolate variables $x$ and $y$ on opposite sides of the equality sign:

$y \, dy = (c - x) \, dx$

Step 3: Integration

Integrate both sides with respect to their respective variables:

$\int y \, dy = \int (c - x) \, dx$

Applying the Power Rule for integration, $\int x^n \, dx = \frac{x^{n+1}}{n+1} + C$:

$\frac{y^2}{2} = cx - \frac{x^2}{2} + K$

where $K$ is the constant of integration. [Per the Fundamental Theorem of Calculus].

Step 4: Algebraic Manipulation to Standard Conic Form

Multiply the entire equation by 2 to clear the denominators:

$y^2 = 2cx - x^2 + 2K$

$x^2 - 2cx + y^2 = 2K$

Complete the square for the $x$ terms:

$(x^2 - 2cx + c^2) + y^2 = 2K + c^2$

$(x - c)^2 + y^2 = (constant)^2$

Let $R^2 = 2K + c^2$. The equation simplifies to:

$(x - c)^2 + (y - 0)^2 = R^2$

Step 5: Geometric Interpretation

The derived equation is in the standard form of a circle $(x - h)^2 + (y - k)^2 = r^2$, where the center is $(h, k)$ and the radius is $r$.

Here, the center of the circle is $(c, 0)$, which lies on the x-axis, and the radius is $R = \sqrt{2K + c^2}$.

Visual Representation

Center (c, 0) x y

Final Solution: The differential equation $y\frac{dy}{dx} + x = c$ represents a family of circles with centers on the x-axis at $(c, 0)$.


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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