Key Concepts and Formulas

• Homogeneous Differential Equations: A differential equation of the form $\frac{dy}{dx} = f(x,y)$ is homogeneous if $f(tx, ty) = f(x,y)$. When $\frac{a}{a'} = \frac{b}{b'}$, we use the substitution $t = ax+by$.
• Separable Differential Equations: A differential equation of the form $g(y) dy = h(x) dx$ is separable. We solve it by integrating both sides: $\int g(y) dy = \int h(x) dx$.
• Integration of Rational Functions: When integrating rational functions, algebraic manipulation (long division or partial fractions) is often necessary to simplify the integrand.

Step-by-Step Solution

Step 1: Identify the type of differential equation and make a substitution.

The given differential equation is $(2x+3y-2)dx + (4x+6y-7)dy = 0$. We can rewrite it as $\frac{dy}{dx} = -\frac{2x+3y-2}{4x+6y-7}$. Since $\frac{2}{4} = \frac{3}{6} = \frac{1}{2}$, the lines are parallel, so we make the substitution $t = 2x+3y$. Why: This substitution simplifies the equation because $2x+3y$ appears in both the numerator and denominator.

Step 2: Rewrite the differential equation in terms of t and x.

From $t = 2x+3y$, we have $3y = t-2x$, so $y = \frac{t-2x}{3}$. Differentiating with respect to $x$, we get $\frac{dy}{dx} = \frac{1}{3} \left(\frac{dt}{dx} - 2\right)$. Also, $4x+6y = 2(2x+3y) = 2t$. Substituting these into the original equation, we have: $\frac{1}{3} \left(\frac{dt}{dx} - 2\right) = -\frac{t-2}{2t-7}$ Why: This step transforms the original differential equation into a new differential equation involving $t$ and $x$.

Step 3: Separate the variables and integrate.

Multiply both sides by 3: $\frac{dt}{dx} - 2 = -\frac{3(t-2)}{2t-7}$ $\frac{dt}{dx} = 2 - \frac{3t-6}{2t-7} = \frac{4t-14-3t+6}{2t-7} = \frac{t-8}{2t-7}$ Separate the variables: $\frac{2t-7}{t-8} dt = dx$ Integrate both sides: $\int \frac{2t-7}{t-8} dt = \int dx$ Why: Separating the variables allows us to integrate each side with respect to its corresponding variable, leading to a solution.

Step 4: Evaluate the integrals.

We have $\int \frac{2t-7}{t-8} dt = \int \frac{2(t-8)+16-7}{t-8} dt = \int \left(2 + \frac{9}{t-8}\right) dt = 2t + 9\ln|t-8| + C_1$. Also, $\int dx = x + C_2$. Therefore, $2t + 9\ln|t-8| = x + C$ where $C = C_2 - C_1$.

Why: Performing the integration is a crucial step to find the general solution.

Step 5: Substitute back and apply the initial condition.

Substitute $t = 2x+3y$ back into the equation: $2(2x+3y) + 9\ln|2x+3y-8| = x + C$ $4x+6y + 9\ln|2x+3y-8| = x + C$ Apply the initial condition $y(0) = 3$. When $x=0$, $y=3$: $4(0) + 6(3) + 9\ln|2(0)+3(3)-8| = 0 + C$ $18 + 9\ln|1| = C$ $C = 18$ Therefore, $4x+6y + 9\ln|2x+3y-8| = x + 18$

Why: The initial condition allows us to find the particular solution by determining the value of the constant of integration.

Step 6: Rewrite the equation in the desired form.

$3x+6y + 9\ln|2x+3y-8| = 18$ Divide by 3: $x+2y + 3\ln|2x+3y-8| = 6$ Comparing with $\alpha x + \beta y + 3\ln|2x+3y-\gamma| = 6$, we have $\alpha = 1$, $\beta = 2$, and $\gamma = 8$.

Why: This step ensures that our solution is in the form required by the problem, which allows us to determine the values of $\alpha$, $\beta$, and $\gamma$.

Step 7: Calculate $\alpha+2\beta+3\gamma$.

$\alpha+2\beta+3\gamma = 1 + 2(2) + 3(8) = 1 + 4 + 24 = 29$

Why: This step calculates the final required value.

Common Mistakes & Tips

• Remember to use absolute value inside the logarithm.
• Carefully perform algebraic manipulations to separate variables and evaluate integrals.
• Double-check the initial condition application and substitution.

Summary

We solved the given differential equation by recognizing it as a type where $\frac{a}{a'} = \frac{b}{b'}$. We used the substitution $t = 2x+3y$ to transform the equation into a separable form, integrated, applied the initial condition, and rewrote the solution in the given form to find $\alpha, \beta, \gamma$. Finally, we calculated $\alpha + 2\beta + 3\gamma$.

The final answer is \boxed{29}.