Question

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

• A. 3
• B. -3
• C. -2
• D. 5

Answer 1

The Correct Option is C

To determine the value of \( a \), given that the solution of the differential equation \(\frac{dy}{dx} = \frac{ax + 3}{2y + 5}\) represents a circle, we first integrate the equation. For the equation to represent a circle, it must match the general form of a circle’s equation: \((x-h)^2 + (y-k)^2 = r^2\). 

Starting with the given equation: \(\frac{dy}{dx} = \frac{ax + 3}{2y + 5}\), separate variables:

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

Now, integrate both sides:

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

This gives:

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

Rearrange it to match the circle equation form:

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

For the equation to represent a circle, the coefficients of \(x^2\) and \(y^2\) should be equal. Thus, equating the coefficients:

\(\frac{-a}{2} = 1\)

Solving for \(a\):

\(-a = 2\)

\(a = -2\)

Therefore, the value of \(a\) is \(-2\), confirming that the solution represents a circle.

Answer 2

Solution: To determine \(a\), solve the given differential equation and check the conditions under which the solution represents a circle.

Rewrite the differential equation: The given equation is:

\[\frac{dy}{dx}=\frac{ax+3}{2y+5}.\]

Separating variables:

\[(2y+5)dy=(ax+3)dx.\]

Integrate both sides: Integrating the left-hand side:

\[\int(2y+5)dy=\int2y~dy+\int5~dy=y^{2}+5y+C_{1},\] where \(C_{1}\) is the constant of integration.

Integrating the right-hand side:

\[\int(ax+3)dx=\int ax~dx+\int3~dx=\frac{ax^{2}}{2}+3x+C_{2},\] where \(C_{2}\) is another constant of integration.

Equating the two sides:

\[y^{2}+5y=\frac{ax^{2}}{2}+3x+C,\] where \(C=C_{2}-C_{1}.\)

Rearrange to standard form: To represent a circle, the equation must take the form:

\[(x-h)^{2}+(y-k)^{2}=r^{2}.\]

The \(y^{2}\) term is already present, but for the \(x^{2}\) term to have the same coefficient as \(y^{2}\), \(a\) must satisfy:

\[\frac{a}{2}=1\Rightarrow a=2.\]
Thus, the value of \(a\) that makes the solution represent a circle is \(a = −2.\)