MHT CET · Maths · Differential Equations
General solution of the differential equation \(\frac{d y}{d x}=\frac{x+y+1}{x+y-1}\) is given by
- A \(x+y=\log |x+y|+c\)
- B \(x-y=\log |x+y|+c\)
- C \(y=x+\log |x+y|+c\)
- D \(y=x \log |x+y|+c\)
Answer & Solution
Correct Answer
(C) \(y=x+\log |x+y|+c\)
Step-by-step Solution
Detailed explanation
\(\frac{d y}{d x}=\frac{x+y+1}{x+y-1}\)
Put \(\quad x+y=t\)
\(\Rightarrow 1+\frac{d y}{d x}=\frac{d t}{d x}\)
\(\Rightarrow \frac{d y}{d x}=\frac{d t}{d x}-1 \)
\( \therefore \frac{d t}{d x}-1=\frac{t+1}{t-1} \)
\( \Rightarrow \frac{d t}{d x}=\frac{t+1+t-1}{t-1} \)
\( \Rightarrow \frac{d t}{d x}=\frac{2 t}{t-1} \)
\( \Rightarrow \left(\frac{t-1}{2 t}\right) d t=d x \)
\( \Rightarrow \left(\frac{1}{2}-\frac{1}{2 t}\right) d t=d x\)
On integrating, we get
\(
\frac{1}{2} t-\frac{1}{2} \log t=x+c_{1}
\)
\(\Rightarrow t-\log t =2 x+2 c_{1} \)
\( \Rightarrow x+y-\log (x+y) =2 x+2 c_{1} \)
\( \Rightarrow y =x+\log (x+y)+c\)
Put \(\quad x+y=t\)
\(\Rightarrow 1+\frac{d y}{d x}=\frac{d t}{d x}\)
\(\Rightarrow \frac{d y}{d x}=\frac{d t}{d x}-1 \)
\( \therefore \frac{d t}{d x}-1=\frac{t+1}{t-1} \)
\( \Rightarrow \frac{d t}{d x}=\frac{t+1+t-1}{t-1} \)
\( \Rightarrow \frac{d t}{d x}=\frac{2 t}{t-1} \)
\( \Rightarrow \left(\frac{t-1}{2 t}\right) d t=d x \)
\( \Rightarrow \left(\frac{1}{2}-\frac{1}{2 t}\right) d t=d x\)
On integrating, we get
\(
\frac{1}{2} t-\frac{1}{2} \log t=x+c_{1}
\)
\(\Rightarrow t-\log t =2 x+2 c_{1} \)
\( \Rightarrow x+y-\log (x+y) =2 x+2 c_{1} \)
\( \Rightarrow y =x+\log (x+y)+c\)
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