If the solution of the differential equation \[ \frac{dy}{dx} = \frac{ax + 3}{2y + 5} \] represents a circle, then $a$ is equal to:

Question

Given that the solution of \[ \frac{d^2y}{dx^2} + \alpha \frac{dy}{dx} + \beta y = -e^{-x} \] is \[ y(x) = C_1 e^{-x} + C_2 e^{2x} + x e^{-x}, \] find the values of $\alpha$ and $\beta$.

Answer 1

To find the value of $ a $ such that the solution to the differential equation $\frac{dy}{dx} = \frac{ax + 3}{2y + 5}$ represents a circle, we first integrate the equation. A circle's general form is $(x-h)^2 + (y-k)^2 = r^2$.

Given the equation: $\frac{dy}{dx} = \frac{ax + 3}{2y + 5}$, separate variables:

$(2y + 5)dy = (ax + 3)dx$

Integrate both sides:

$\int (2y + 5)dy = \int (ax + 3)dx$

This yields:

$y^2 + 5y = \frac{ax^2}{2} + 3x + C$

Rearrange to resemble the circle equation:

$y^2 + 5y - \left(\frac{ax^2}{2} + 3x + C\right) = 0$

For this equation to be a circle, the coefficients of $x^2$ and $y^2$ must be equal. Equating these coefficients:

$\frac{-a}{2} = 1$

Solve for $a$:

$-a = 2$

$a = -2$

Thus, $a = -2$ ensures the solution represents a circle.

If the solution of the differential equation \[ \frac{dy}{dx} = \frac{ax + 3}{2y + 5} \] represents a circle, then $a$ is equal to:

The Correct Option is C

Solution and Explanation

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