TS EAMCET · Maths · Differential Equations
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 \(5 \mathrm{x}-15 \mathrm{y}-4 \log |15 \mathrm{x}-20 \mathrm{y}-12|=\mathrm{c}\)
- C \(5 \mathrm{x}-15 \mathrm{y}+14 \log |15 \mathrm{x}-20 \mathrm{y}-12|=\mathrm{c}\)
- D \(8 y-4 x+\log |9 x-12 y+4|=c\)
Answer & Solution
Correct Answer
(C) \(5 \mathrm{x}-15 \mathrm{y}+14 \log |15 \mathrm{x}-20 \mathrm{y}-12|=\mathrm{c}\)
Step-by-step Solution
Detailed explanation
\[ \begin{aligned} & \text { }(3 x-4 y)(d x-3 d y)+(6 d x-4 d y)=0 \\ & \Rightarrow \frac{d y}{d x}=\frac{(3 x-4 y)+6}{3(3 x-4 y)+4} \end{aligned} \] Let \(3 x-4 y=t \Rightarrow 3-4 \frac{d y}{d x}=\frac{d t}{d x}\)…
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