MHT CET · Maths · Differential Equations
The solution of the equation \(\frac{\mathrm{dy}}{\mathrm{d} x}=\frac{1}{x+\mathrm{y}+1}\) is
- A \(x=\log (x+\mathrm{y}+2)+\mathrm{c}\), where c is the constant of integration
- B \(x=\log (x+\mathrm{y}-2)+\mathrm{c}\), where c is the constant of integration
- C \(\mathrm{y}=\log (x+\mathrm{y}+2)+\mathrm{c}\), where c is the constant of integration
- D \(\mathrm{y}=\log (x+\mathrm{y}-2)+\mathrm{c}\), where c is the constant of integration
Answer & Solution
Correct Answer
(C) \(\mathrm{y}=\log (x+\mathrm{y}+2)+\mathrm{c}\), where c is the constant of integration
Step-by-step Solution
Detailed explanation
Let \( v = x+y+1 \). Then \( \frac{\mathrm{d}v}{\mathrm{d}x} = 1 + \frac{\mathrm{dy}}{\mathrm{d}x} \implies \frac{\mathrm{dy}}{\mathrm{d}x} = \frac{\mathrm{d}v}{\mathrm{d}x} - 1 \). Substitute into the equation: \( \frac{\mathrm{d}v}{\mathrm{d}x} - 1 = \frac{1}{v} \).
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