MHT CET · Maths · Differential Equations
The particular solution of the differential equation \(\frac{d y}{d x}=\frac{y+1}{x^2-x}\), when \(\mathrm{x}=2\) and \(\mathrm{y}=1\) is
- A \(x y=4 x-6\)
- B xy \(=2 x-2\)
- C \(x y=x-2\)
- D \(x y=-x+4\)
Answer & Solution
Correct Answer
(C) \(x y=x-2\)
Step-by-step Solution
Detailed explanation
\(\frac{d y}{d x}=\frac{y+1}{x^2-x} \)
\( \therefore \int \frac{d y}{y+1}=\int \frac{d x}{x(x-1)} \)
\( \therefore \int \frac{d y}{y+1}=\int\left[\frac{1}{x-1}-\frac{1}{x}\right] d x \Rightarrow \log |y+1|=\) \(\log |x-1|-\log |x|+ \)
\( \log c \)
\( \text { We have } x=2 \text { and } y=1 \)
\( \therefore \log |2|=\log |1|-\log |1|+\log c \Rightarrow c=2 \)
\( \therefore \log |y+1|=\log |x-1|-\log |x|+\log |2| \)
\( \log |y+1|=\log \left|\frac{2(x-1)}{x}\right| \)
\( \therefore y+1=\frac{2(x-1)}{x} \Rightarrow x y+x=2 x-2 \Rightarrow x y=x-2\)
\( \therefore \int \frac{d y}{y+1}=\int \frac{d x}{x(x-1)} \)
\( \therefore \int \frac{d y}{y+1}=\int\left[\frac{1}{x-1}-\frac{1}{x}\right] d x \Rightarrow \log |y+1|=\) \(\log |x-1|-\log |x|+ \)
\( \log c \)
\( \text { We have } x=2 \text { and } y=1 \)
\( \therefore \log |2|=\log |1|-\log |1|+\log c \Rightarrow c=2 \)
\( \therefore \log |y+1|=\log |x-1|-\log |x|+\log |2| \)
\( \log |y+1|=\log \left|\frac{2(x-1)}{x}\right| \)
\( \therefore y+1=\frac{2(x-1)}{x} \Rightarrow x y+x=2 x-2 \Rightarrow x y=x-2\)
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