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MHT CET · Maths · Differential Equations

The particular solution of the differential equation \(\left(y+x \cdot \frac{d y}{d x}\right) \cdot \sin x y=\cos x\)
at \(x=0\) is

  1. A \(\sin x+\cos x y=1\)
  2. B \(\cos x-\sin x y=1\)
  3. C \(\sin x-\cos x y=1\)
  4. D \(\cos x+\sin x y=1\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(\sin x+\cos x y=1\)

Step-by-step Solution

Detailed explanation

We have \(\left(y+x \frac{d y}{d x}\right) \sin x y=\cos x\)
Put \(x y=u \Rightarrow x \frac{d y}{d x}+y=\frac{d u}{d x}\)
\(\therefore\left(\frac{\mathrm{du}}{\mathrm{dx}}\right) \sin \mathrm{u}=\cos \mathrm{x}\)
\(\therefore \int \sin u d u=\int \cos x d x \Rightarrow-\cos u=\sin x+c \Rightarrow\) \(-\cos x y=\sin x+c\)
When \(x=0\), we get
\(-\cos 0=0+c \Rightarrow c=-1\)
\(\therefore-\cos x y=\sin x-1 \Rightarrow \sin x+\cos x y=1\)
From MHT CET
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