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MHT CET · Maths · Differential Equations

The solution of the differential equation \(\sin ^{-1}\left(\frac{\mathrm{dy}}{\mathrm{d} x}\right)=x+\mathrm{y}\) is

  1. A \(x=\tan (x+y) \cdot \sec (x+y)+c\)
  2. B \(x=\tan (x+y)-\sec (x+y)+c\)
  3. C \(x=\tan (x+y)+\sec (x+y)+c\)
  4. D \(x=\tan x \cdot \tan y+c\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(x=\tan (x+y)-\sec (x+y)+c\)

Step-by-step Solution

Detailed explanation

We have \(\sin ^{-1}\left(\frac{d y}{d x}\right)=x+y\)
\(
\therefore \frac{\mathrm{dy}}{\mathrm{dx}}=\sin (\mathrm{x}+\mathrm{y})
\)
Put \(\quad x+y=t \Rightarrow y=t-x \Rightarrow \frac{d y}{d x}=\frac{d t}{d x}-1\)
\(
\therefore \frac{\mathrm{dt}}{\mathrm{dx}}=1+\sin \mathrm{t} \Rightarrow \int \frac{\mathrm{dt}}{1+\sin \mathrm{t}}=\int \mathrm{dx}
\)
\(\int \frac{(1-\sin t)}{(1+\sin t)(1-\sin t)} d t=\int d x \Rightarrow x=\int \frac{1-\sin t}{1-\sin ^{2} t} d t\)
\(x=\int \frac{1-\sin t}{\cos ^{2} t} d t=\int\left(\sec ^{2} t-\sec t \tan t\right) d t\)
\(x=\tan t-\sec t+c\)
\(x=\tan (x+y)-\sec (x+y)+c\)