If the solution of the differential equation $(2x+3y-<!--\; -\; -->2)dx+(4x+6y-<!--\; -\; -->7)dy=0,y(0)=3$, is $\mathrm{\alpha <!--\; \alpha \; -->}x+\mathrm{\beta <!--\; \beta \; -->}y+3{\log }_e<!--\; \; -->|2x+3y-<!--\; -\; -->\mathrm{\gamma <!--\; \gamma \; -->}|=6$, then $\mathrm{\alpha <!--\; \alpha \; -->}+2\mathrm{\beta <!--\; \beta \; -->}+3\mathrm{\gamma <!--\; \gamma \; -->}$ is equal to ____

The correct answer is 29.

We are given the differential equation:

$(2x+3y-2)dx+(4x+6y-7)dy=0$, with $y(0)=3$

Step 1: Identify the structure.

Notice that the coefficient of $dy$ is $4x+6y-7=2(2x+3y)-7$. Both coefficients share the combination $2x+3y$, suggesting the substitution $v=2x+3y$.

Step 2: Apply the substitution $v=2x+3y$.

Differentiating with respect to $x$:

$\frac{d}{dx}v=2+3\frac{d}{y}$

Rewrite the ODE as:

$\frac{d}{y}=-\frac{2x+3y-2}{4x+6y-7}=-\frac{v-2}{2v-7}$

Substituting into the expression for $\frac{d}{v}$:

$\frac{d}{v}=2+3·(-\frac{v-2}{2v-7})=\frac{2(2v-7)-3(v-2)}{2v-7}=\frac{v-8}{2v-7}$

Step 3: Separate variables and integrate.

We get:

$\frac{2v-7}{v-8}dv=dx$

Perform polynomial long division on the left side. Since $2v-7=2(v-8)+9$:

$\frac{2v-7}{v-8}=2+\frac{9}{v-8}$

Integrating both sides:

$\int (2+\frac{9}{v-8})dv=\int dx$

$2v+9{\log }_e|v-8|=x+C$

Step 4: Substitute back $v=2x+3y$.

$2(2x+3y)+9{\log }_e|2x+3y-8|=x+C$

$4x+6y+9{\log }_e|2x+3y-8|=x+C$

Rearranging:

$3x+6y+9{\log }_e|2x+3y-8|=C$

Dividing through by 3:

$x+2y+3{\log }_e|2x+3y-8|=\frac{C}{3}$

Step 5: Apply the initial condition $y(0)=3$.

At $x=0$, $y=3$:

$0+2(3)+3{\log }_e|2(0)+3(3)-8|=\frac{C}{3}$

$6+3{\log }_e|9-8|=\frac{C}{3}$

$6+3{\log }_{e}1=\frac{C}{3}\Rightarrow \frac{C}{3}=6$

So the particular solution is:

$x+2y+3{\log }_e|2x+3y-8|=6$

Step 6: Identify α, β, and γ by comparison.

The given form of the solution is $\alpha x+\beta y+3{\log }_e|2x+3y-\gamma |=6$.

Comparing with our derived solution $x+2y+3{\log }_e|2x+3y-8|=6$, we identify:

$\alpha =1$, $\beta =2$, $\gamma =8$

Step 7: Calculate the required expression.

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