If the solution of the differential equation
$(2 x+3 y-2) \mathrm{d} x+(4 x+6 y-7) \mathrm{d} y=0, y(0)=3$, is
$\alpha x+\beta y+3 \log _e|2 x+3 y-\gamma|=6$, then $\alpha+2 \beta+3 \gamma$ is equal to ____________.
Answer (integer)
29
Solution
<p>$$\begin{array}{ll}
2 x+3 y-2=t & 4 x+6 y-4=2 t \\
2+3 \frac{d y}{d x}=\frac{d t}{d x} & 4 x+6 y-7=2 t-3
\end{array}$$</p>
<p>$$\begin{aligned}
& \frac{d y}{d x}=\frac{-(2 x+3 y-2)}{4 x+6 y-7} \\
& \frac{d t}{d x}=\frac{-3 t+4 t-6}{2 t-3}=\frac{t-6}{2 t-3} \\
& \int \frac{2 t-3}{t-6} d t=\int d x \\
& \int\left(\frac{2 t-12}{t-6}+\frac{9}{t-6}\right) \cdot d t=x \\
& 2 t+9 \ln (t-6)=x+c \\
& 2(2 x+3 y-2)+9 \ln (2 x+3 y-8)=x+c \\
& x=0, y=3 \\
& c=14 \\
& 4 x+6 y-4+9 \ln (2 x+3 y-8)=x+14 \\
& x+2 y+3 \ln (2 x+3 y-8)=6 \\
& \alpha=1, \beta=2, \gamma=8 \\
& \alpha+2 \beta+3 \gamma=1+4+24=29
\end{aligned}$$</p>
About this question
Subject: Mathematics · Chapter: Differential Equations · Topic: Order and Degree
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