Q119 / 160 · TS EAMCET 2023 (13 May Shift 2) · Mathematics

The general solution of the differential equation $(3 x-4 y)$ $(d x-3 d y)+(6 d x-4 d y)=0$ is

  • A
    $x-2 y+\log |3 x-4 y+6|=c$
  • B
    $5 \mathrm{x}-15 \mathrm{y}-4 \log |15 \mathrm{x}-20 \mathrm{y}-12|=\mathrm{c}$
  • Correct answer
    C
    $5 \mathrm{x}-15 \mathrm{y}+14 \log |15 \mathrm{x}-20 \mathrm{y}-12|=\mathrm{c}$
  • D
    $8 y-4 x+\log |9 x-12 y+4|=c$

Solution

$$ \begin{aligned} & \text { }(3 x-4 y)(d x-3 d y)+(6 d x-4 d y)=0 \\ & \Rightarrow \frac{d y}{d x}=\frac{(3 x-4 y)+6}{3(3 x-4 y)+4} \end{aligned} $$ Let $3 x-4 y=t \Rightarrow 3-4 \frac{d y}{d x}=\frac{d t}{d x}$ $$ \begin{aligned} & \Rightarrow \quad \frac{1}{4}\left[3-\frac{d t}{d x}\right]=\frac{t+6}{3 t+4} \Rightarrow \frac{d t}{d x}=3-\frac{4(t+6)}{(3 t+4)} \\ & \Rightarrow \quad \frac{d t}{d x}=\frac{9 t+12-4 t-24}{(3 t+4)} \\ & \Rightarrow \quad \frac{d t}{d x}=\frac{5 t-12}{3 t+4} \Rightarrow d x=\left(\frac{3 t+4}{5 t-12}\right) d t \\ & \Rightarrow \quad d x=\left[\frac{3}{5}\left(\frac{5 t-12}{5 t-12}\right)+\frac{56}{5(5 t-12)}\right] d t \\ & \Rightarrow \quad x=\frac{3}{5} \int \frac{5 t-12}{5 t-12} d t+\frac{56}{5} \int \frac{d t}{5 t-12} \\ & \Rightarrow \quad x=\frac{3}{5} t+\frac{56}{25} \log |5 t-12|+C \\ & \Rightarrow \quad x=\frac{3}{5}(3 x-4 y)+\frac{56}{25} \log |15 x-20 y+12|+C \\ & \Rightarrow \quad 5 x-15 y+14 \log |15 x-20 y+12|=C \end{aligned} $$

Hence, the correct answer is option: 'C'.

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