If the solution of the differential equation (2x + 3y - 2)dx + (4x + 6y - 7)dy = 0, y(0) = 3, is αx

Q: Wait, why can't we just use the standard linear differential equation method here?

That is a great question! The standard linear form $\frac{dy}{dx} + Py = Q$ requires $P$ and $Q$ to be functions of $x$ only. In our equation, the coefficients of $dx$ and $dy$ are both functions of $x$ and $y$. Because the lines $2x + 3y - 2 = 0$ and $4x + 6y - 7 = 0$ are parallel, we cannot use the intersection point method, but we can use the substitution method to reduce it to a separable form.

Q: Why did we choose $v = 2x + 3y$ specifically?

We chose $v = 2x + 3y$ because it is the common linear factor appearing in both parts of the differential equation. By setting $v$ equal to this expression, we effectively reduce the two-dimensional problem into a one-dimensional one, making the equation separable. It is the most efficient way to handle parallel linear terms.

Q: What if the lines were not parallel? How would the approach change?

If the lines were not parallel, they would intersect at a unique point $(h, k)$. In that case, we would shift the origin to $(h, k)$ by substituting $x = X + h$ and $y = Y + k$. This would eliminate the constant terms, leaving us with a homogeneous differential equation, which we would then solve by substituting $Y = vX$.

Question

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

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on a vast, flat coordinate plane. You are presented with a differential equation:
$$(2x + 3y - 2)dx + (4x + 6y - 7)dy = 0$$

At first glance, it looks like a tangled mess of variables. However, look at the linear expressions $2x + 3y - 2$ and $4x + 6y - 7$.

The coefficients of $x$ and $y$ in the second expression are exactly double those in the first. Geometrically, these represent two parallel lines. This is a structural invitation to simplify.

### The Power of Substitution

Since the lines are parallel, we can exploit this symmetry. We define a new variable $v = 2x + 3y$.

By bundling this repeating pattern into a single variable, we are essentially changing our coordinate system to align with the geometry of the problem. Differentiating $v$ with respect to $x$ gives:
$$\frac{dv}{dx} = 2 + 3\frac{dy}{dx}$$

Rearranging this, we find that:
$$\frac{dy}{dx} = \frac{1}{3} \left( \frac{dv}{dx} - 2 \right)$$
This serves as our bridge between the old world of $x$ and $y$ and the new, simpler world of $v$ and $x$.

### The Transformation

Now, we rewrite our original differential equation in terms of $\frac{dy}{dx}$:
$$\frac{dy}{dx} = -\frac{2x + 3y - 2}{4x + 6y - 7}$$

Substituting our new variable $v$, the equation becomes:
$$\frac{1}{3} \left( \frac{dv}{dx} - 2 \right) = -\frac{v - 2}{2v - 7}$$

The variables $x$ and $y$ have vanished, leaving us with a separable differential equation. After algebraic manipulation, we arrive at:
$$\frac{dv}{dx} = \frac{v - 8}{2v - 7}$$

### The Art of Integration

With the variables separated, we have:
$$\int \frac{2v - 7}{v - 8} dv = \int dx$$

To integrate the left side, we use an adjustment trick. We rewrite the numerator to match the denominator:
$$\frac{2(v - 8) + 9}{v - 8} = 2 + \frac{9}{v - 8}$$

Now, the integration is straightforward:
$$\int \left( 2 + \frac{9}{v - 8} \right) dv = \int dx$$
This yields the result $2v + 9 \ln |v - 8| = x + C$.

### The Final Polish

Finally, we substitute $v = 2x + 3y$ back into our solution to return to the original variables:
$$2(2x + 3y) + 9 \ln |2x + 3y - 8| = x + C$$

Simplifying this, we get:
$$3x + 6y + 9 \ln |2x + 3y - 8| = C$$

To match the target format $\alpha x + \beta y + 3 \ln |2x + 3y - \gamma| = 6$, we divide the entire equation by $3$:
$$x + 2y + 3 \ln |2x + 3y - 8| = C'$$

Using the initial condition $y(0) = 3$, we find $C' = 6$. By direct comparison, $\alpha = 1$, $\beta = 2$, and $\gamma = 8$.

The final calculation is:
$$\alpha + 2\beta + 3\gamma = 1 + 4 + 24 = \mathbf{29}$$

Animated Solution for Mathematics - Differential Equations: If the solution of the differential equation $(2 x + 3 y - 2) d x + (4 x + 6 y - 7) d y = 0, y (0) = 3$ , is $α x + β y + 3 lo g_{e} ∣2 x + 3 y - γ ∣ = 6$ , then $α + 2 β + 3 γ$ is equal to_____

Enter Numerical Value:

Visualized Solution

Analyze the Equation Structure

Substitution $v = 2 x + 3 y$

Differentiating $v$ with respect to $x$

Rewriting the Differential Equation

Substituting $v$ into the Equation

Algebraic Simplification

Variable Separation

Preparing for Integration

Performing Integration

Back Substitution and Simplification

Matching the Target Format

Applying Initial Condition $y (0) = 3$

Comparing and Finding $α, β, γ$

Final Calculation

The Sigma Insight: Variable Separable Method

Analyzing the Setup

The Power of Substitution

The Transformation

The Art of Integration

The Final Polish

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Enter Numerical Value:

Visualized Solution

Analyze the Equation Structure

Substitution .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } v=2x+3y

Differentiating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } v with respect to .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x