Question

If the solution of the differential equation $(2x + 3y - 2) \, dx + (4x + 6y - 7) \, dy = 0, \quad y(0) = 3,$ is $\alpha x + \beta y + 3 \log_e |2x + 3y - \gamma| = 6,$ then $\alpha + 2\beta + 3\gamma$ is equal to ____.

Answer 1

Given the differential equation:

$(2x + 3y - 2)dx + (4x + 6y - 7)dy = 0, \quad y(0) = 3$
We define:
$t = 2x + 3y - 2$
Differentiating with respect to $x$:

$\frac{dt}{dx} = 2 + 3 \frac{dy}{dx}$

Rearranging:

$\frac{dy}{dx} = \frac{\frac{dt}{dx} - 2}{3}$

Step 1. Substituting into the Original Equation: Substituting $\frac{dy}{dx}$ into the given differential equation:

$(2x + 3y - 2)dx + (4x + 6y - 7) \left( \frac{\frac{dt}{dx} - 2}{3} \right) dx = 0$

Step 2. Simplifying:

$3(2x + 3y - 2) + (4x + 6y - 7) \left( \frac{dt}{dx} - 2 \right) = 0$

Further simplification leads to separation of terms and integration.
Integrating Both Sides: Integrating both sides with respect to $x$ yields:

$\int ...$

Step 3. Solving for Constants: Given the initial condition $y(0) = 3$, we can find the value of constants.

Step 4. Finding the Value of $\alpha, \beta, \gamma$**: Substituting known values, we find:

$\alpha + 2\beta + 3\gamma = 29$