Differential Equations Question 6
Question 6 - 2024 (27 Jan Shift 1)
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
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Answer (29)
Solution
$$ \begin{aligned} & 2 x+3 y-2=t \quad 4 x+6 y-4=2 t \\ & 2+3 \frac{d y}{d x}=\frac{d t}{d x} \quad 4 x+6 y-7=2 t-3 \\ & \frac{d y}{d x}=\frac{-(2 x+3 y-2)}{4 x+6 y-7} \\ & \frac{d t}{d x}=\frac{-3 t+4 t-6}{2 t-3}=\frac{t-6}{2 t-3} \\ & \int \frac{2 t-3}{t-6} d t=\int d x \\ & \int\left(\frac{2 t-12}{t-6}+\frac{9}{t-6}\right) \cdot d t=x \\ & 2 t+9 \ln (t-6)=x+c \\ & 2(2 x+3 y-2)+9 \ln (2 x+3 y-8)=x+c \\ & x=0, y=3 \\ & c=14 \\ & 4 x+6 y-4+9 \ln (2 x+3 y-8)=x+14 \\ & x+2 y+3 \ln (2 x+3 y-8)=6 \\ & \alpha=1, \beta=2, \gamma=8 \\ & \alpha+2 \beta+3 \gamma=1+4+24=29 \end{aligned} $$
