MHT CET · Maths · Differential Equations
The integrating factor of the differential equation siny \(\left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)=\operatorname{cosy}(1-x \cos y)\) is
- A \(\mathrm{e}^{-x}\)
- B \(\mathrm{e}^{-\cos y}\)
- C \(\mathrm{e}^{-y}\)
- D \(\mathrm{e}^{\text {siny }}\)
Answer & Solution
Correct Answer
(A) \(\mathrm{e}^{-x}\)
Step-by-step Solution
Detailed explanation
(A)
\(\sin y \frac{d y}{d x}=\cos y(1-x \cos y) \Rightarrow \sin y \frac{d y}{d x}=\) \(\cos y-x \cos ^{2} y\)
Dividing both sides by \(\cos ^{2} y\), we get
\(\sec y \tan y \frac{d y}{d x}=\sec y-x\)
\(\sec y \cdot \tan y \frac{d y}{d x}-\sec y=-x\)
Put \(\sec y=v \Rightarrow \sec y \tan y d y=d x\)
\(\therefore \frac{\mathrm{dv}}{\mathrm{d} \mathrm{x}}-\mathrm{v}=-\mathrm{x}\)...(1)
\(\therefore\) I.F. \(\quad=\mathrm{e}^{-\int 1 \mathrm{dx}}=\mathrm{e}^{-\mathrm{x}}\)
\(\sin y \frac{d y}{d x}=\cos y(1-x \cos y) \Rightarrow \sin y \frac{d y}{d x}=\) \(\cos y-x \cos ^{2} y\)
Dividing both sides by \(\cos ^{2} y\), we get
\(\sec y \tan y \frac{d y}{d x}=\sec y-x\)
\(\sec y \cdot \tan y \frac{d y}{d x}-\sec y=-x\)
Put \(\sec y=v \Rightarrow \sec y \tan y d y=d x\)
\(\therefore \frac{\mathrm{dv}}{\mathrm{d} \mathrm{x}}-\mathrm{v}=-\mathrm{x}\)...(1)
\(\therefore\) I.F. \(\quad=\mathrm{e}^{-\int 1 \mathrm{dx}}=\mathrm{e}^{-\mathrm{x}}\)
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