Explanation
\begin{aligned}
& \text{Given:D.E.is} \\
& (2x+3y-2)dx+(4x+6y-7)dy=0 \\
& \Rightarrow\frac{dy}{dx}=\frac{-2x+3y-2}{4x+6y-7} & ...(\mathbf{i}) \\
& \operatorname{Let}2x+3y=\mu \\
& \Rightarrow2+3\frac{dy}{dx}=\frac{du}{dx} \\
& \text{So, (i) becomes} \\
& \Rightarrow\frac{1}{3}\left(\frac{du}{dx}-2\right)=-\frac{u-2}{24-7} \\
& \mathrm{=}\Rightarrow{\frac{du}{dx}}={\frac{-3u+6}{24-7}}+2={\frac{-3u+6+4u-14}{2u-7}} \\
& \Rightarrow\frac{du}{dx}=\frac{4-8}{24-7} \\
& \Rightarrow\int\frac{24-7}{4-8}du=\int dx \\
& \Rightarrow\int\left[1+\frac{4+1}{4-8}\right]du=\int dx
\end{aligned}
2u+ln∣4−8∣=x+C
2(2x+3y)+9ln∣2x+3y−8∣=x+C
Given: u(0,0)=3
2(2(0)+3(0))+9+9ln∣2(0)+4−8∣=0+C
18+9(0)=C⇒C=18
Solution is:
4x+6y+9ln∣2x+3y−8∣=x+18
3x+6y+9ln∣2x+3y−8∣=18
x+2y+3ln∣2x+3y−8∣=6
On comparing, we get α=1, β=2, γ=8
∴ α+2β+3γ=1+4+24=29