Question

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 \(\alpha x+\beta y+3 \log _{e}|2 x+3 y-\gamma|=6\), then \(\alpha+2 \beta+3 \gamma\) is equal to ____.

Answer 1

Correct answer: 29

\(2 \mathrm{x}+3 \mathrm{y}-2=\mathrm{t}\)

\(2+3 \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{dt}}{\mathrm{dx}}\)

\(4 \mathrm{x}+6 \mathrm{y}-4=2 \mathrm{t}\)

\(4 \mathrm{x}+6 \mathrm{y}-7=2 \mathrm{t}-3\)

\(\frac{d y}{d x}=\frac{-(2 x+3 y-2)}{4 x+6 y-7}\)

\(\frac{\mathrm{dt}}{\mathrm{dx}}=\frac{-3 \mathrm{t}+4 \mathrm{t}-6}{2 \mathrm{t}-3}=\frac{\mathrm{t}-6}{2 \mathrm{t}-3}\)

\(\int \frac{2 \mathrm{t}-3}{\mathrm{t}-6} \mathrm{dt}=\int \mathrm{dx}\)

\(\int\left(\frac{2 t-12}{t-6}+\frac{9}{t-6}\right) \cdot d t=x\)

\(2 \mathrm{t}+9 \ln (\mathrm{t}-6)=\mathrm{x}+\mathrm{c}\)

\(2(2 x+3 y-2)+9 \ln (2 x+3 y-8)=x+c\)

\(\mathrm{x}=0, \mathrm{y}=3\)

\(\mathrm{c}=14\)

\(4 \mathrm{x}+6 \mathrm{y}-4+9 \ln (2 \mathrm{x}+3 \mathrm{y}-8)=\mathrm{x}+14\)

\(\mathrm{x}+2 \mathrm{y}+3 \ln (2 \mathrm{x}+3 \mathrm{y}-8)=6\)

\(\alpha=1, \beta=2, \gamma=8\)

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