Question

If the solution of the differential equation \(\frac{dy}{dx}=\frac{2x+3y}{3x-2y}\) is y = xtan(f(x)) + c then f(x) =

• A. \(\frac13log(x^2+y^2)\)
• B. (2x+3y)logx
• C. xlog\(\frac{y}{x}\)+y²
• D. sin(x+y²)

Answer 1

The Correct Option is A

To solve the problem, we need to find the general solution of the homogeneous differential equation $\frac{dy}{dx} = \frac{2x + 3y}{3x - 2y}$ and identify the coefficient of $\log(x^2 + y^2)$ in the expression for $\arctan\left( \frac{y}{x} \right)$.

1. Verify Homogeneity:
The equation is homogeneous since $\frac{2x + 3y}{3x - 2y} = \frac{2 + 3(y/x)}{3 - 2(y/x)}$.
Let $y = vx$, so $\frac{dy}{dx} = v + x \frac{dv}{dx}$.

2. Substitute and Simplify:
Substitute into the equation:
$v + x \frac{dv}{dx} = \frac{2 + 3v}{3 - 2v}$.
Solve for $x \frac{dv}{dx}$:
$x \frac{dv}{dx} = \frac{2 + 3v}{3 - 2v} - v = \frac{2 + 3v - 3v + 2v^2}{3 - 2v} = \frac{2 + 2v^2}{3 - 2v}$.
Thus:
$\frac{3 - 2v}{2(1 + v^2)} dv = \frac{dx}{x}$.

3. Integrate Both Sides:
Left-hand side:
$\int \frac{3 - 2v}{2(1 + v^2)} dv = \int \left( \frac{3}{2(1 + v^2)} - \frac{v}{1 + v^2} \right) dv = \frac{3}{2} \arctan v - \frac{1}{2} \log(1 + v^2) + C_1$.
Right-hand side:
$\int \frac{dx}{x} = \log |x| + C_2$.
Equate:
$\frac{3}{2} \arctan v - \frac{1}{2} \log(1 + v^2) = \log |x| + C$.

4. Substitute Back:
Since $v = \frac{y}{x}$:
$\frac{3}{2} \arctan\left( \frac{y}{x} \right) - \frac{1}{2} \log\left( 1 + \frac{y^2}{x^2} \right) = \log |x| + C$.
Simplify:
$\frac{3}{2} \arctan\left( \frac{y}{x} \right) - \frac{1}{2} \log(x^2 + y^2) + \log |x| = \log |x| + C$.
Thus:
$\frac{3}{2} \arctan\left( \frac{y}{x} \right) - \frac{1}{2} \log(x^2 + y^2) = C$.

5. Simplify the Solution:
Multiply by 2:
$3 \arctan\left( \frac{y}{x} \right) - \log(x^2 + y^2) = k$.
Rewrite:
$\arctan\left( \frac{y}{x} \right) = \frac{1}{3} \log(x^2 + y^2) + \frac{k}{3}$.
The coefficient of $\log(x^2 + y^2)$ is $\frac{1}{3}$.

Final Answer:
The coefficient of $\log(x^2 + y^2)$ is $\frac{1}{3}$.

Answer 2

Step 1: Verify Homogeneity

The given equation is homogeneous because it can be rewritten as:

(2x + 3y) / (3x - 2y) = (2 + 3(y/x)) / (3 - 2(y/x))

Now, let’s assume that y = vx, where v = y/x. Thus, dy/dx = v + x(dv/dx).

Step 2: Substitute and Simplify

Substitute y = vx and dy/dx = v + x(dv/dx) into the equation:

v + x(dv/dx) = (2 + 3v) / (3 - 2v)

Now, solve for x(dv/dx):

x(dv/dx) = (2 + 3v) / (3 - 2v) - v = (2 + 3v - 3v + 2v²) / (3 - 2v) = (2 + 2v²) / (3 - 2v)

Thus, the equation becomes:

(3 - 2v) / (2(1 + v²)) dv = dx / x

Step 3: Integrate Both Sides

Now, we will integrate both sides of the equation:

Left-hand side:

        ∫ (3 - 2v) / (2(1 + v²)) dv        = ∫ (3 / 2(1 + v²)) - (v / (1 + v²)) dv        = (3/2) arctan(v) - (1/2) log(1 + v²) + C₁    

Right-hand side:

        ∫ (dx / x) = log |x| + C₂    

Equating both sides:

        (3 / 2) arctan(v) - (1 / 2) log(1 + v²) = log |x| + C    

Step 4: Substitute Back

Recall that v = y/x. Substitute this back into the equation:

        (3 / 2) arctan(y/x) - (1 / 2) log(1 + (y²/x²)) = log |x| + C    

Now simplify the expression:

        (3 / 2) arctan(y/x) - (1 / 2) log(x² + y²) + log |x| = log |x| + C    

Thus, we have:

        (3 / 2) arctan(y/x) - (1 / 2) log(x² + y²) = C    

Step 5: Simplify the Final Solution

To simplify further, multiply through by 2:

        3 arctan(y/x) - log(x² + y²) = k    

Rearranging the terms gives us:

        arctan(y/x) = (1/3) log(x² + y²) + k/3    

Final Answer:

The coefficient of log(x² + y²) is 1/3.