MHT CET · Maths · Differential Equations
\(If \frac{x}{x-y}=\log \left(\frac{a}{x-y}\right), \text { then } \frac{d y}{d x}=\)
- A \(2+\frac{1}{y}\)
- B \(\frac{2 y-x}{y}\)
- C \(\frac{2 x-y}{x}\)
- D \(\frac{x-2 y}{y}\)
Answer & Solution
Correct Answer
(B) \(\frac{2 y-x}{y}\)
Step-by-step Solution
Detailed explanation
\(\frac{x}{x-y}=\log a-\log (x-y)\)
\(\therefore \log (x-y)+\frac{x}{x-y}=\log a\)
\(\therefore \frac{1}{(x-y)}\left(1-\frac{d y}{d x}\right)+\left[\frac{(x-y)(1)-(x)\left(1-\frac{d y}{d x}\right)}{x-y}\right]=0\)
\(\therefore\left[\frac{1}{x-y}-\frac{1}{x-y} \frac{d y}{d x}\right]+\left[\frac{x-y-x+x \frac{d y}{d x}}{x-y}\right]=0\)
\(\therefore \frac{1}{x-y}-\frac{1}{x-y} \frac{d y}{d x}-\frac{y}{(x-y)^{2}}+\frac{x}{(x-y)^{2}}=0\)
\(\therefore \quad(x-y)-(x-y) \frac{d y}{d x}-y+x \frac{d y}{d x}=0\)
\(\therefore \frac{d y}{d x}=\frac{x-2 y}{-y}=\frac{2 y-x}{y}\)
\(\therefore \log (x-y)+\frac{x}{x-y}=\log a\)
\(\therefore \frac{1}{(x-y)}\left(1-\frac{d y}{d x}\right)+\left[\frac{(x-y)(1)-(x)\left(1-\frac{d y}{d x}\right)}{x-y}\right]=0\)
\(\therefore\left[\frac{1}{x-y}-\frac{1}{x-y} \frac{d y}{d x}\right]+\left[\frac{x-y-x+x \frac{d y}{d x}}{x-y}\right]=0\)
\(\therefore \frac{1}{x-y}-\frac{1}{x-y} \frac{d y}{d x}-\frac{y}{(x-y)^{2}}+\frac{x}{(x-y)^{2}}=0\)
\(\therefore \quad(x-y)-(x-y) \frac{d y}{d x}-y+x \frac{d y}{d x}=0\)
\(\therefore \frac{d y}{d x}=\frac{x-2 y}{-y}=\frac{2 y-x}{y}\)
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