MHT CET · Maths · Differential Equations
The solution of the differential equation \(\mathrm{e}^{-x}(y+1) \mathrm{d} y+\left(\cos ^2 x-\sin 2 x\right) y \mathrm{~d} x=0\) at \(x=0\), \(y=1\) is
- A \((y+1)+\mathrm{e}^x \cos ^2 x=2\)
- B \(y+\log y=\mathrm{e}^x \cos ^2 x\)
- C \(\log (y+1)+\mathrm{e}^x \cos ^2 x=1\)
- D \(y+\log y+\mathrm{e}^x \cos ^2 x=2\)
Answer & Solution
Correct Answer
(D) \(y+\log y+\mathrm{e}^x \cos ^2 x=2\)
Step-by-step Solution
Detailed explanation
\( \mathrm{e}^{-x}(y+1) \mathrm{d} y+\left(\cos ^2 x-\sin 2 x\right) y \mathrm{~d} x=0 \)
\( \Rightarrow \frac{y+1}{y} \mathrm{~d} y+\mathrm{e}^x\left(\cos ^2 x-\sin 2 x\right) \mathrm{d} x=0\)
Integrating on both sides, we get
\( \int\left(1+\frac{1}{y}\right) \mathrm{d} y+\int \mathrm{e}^x\left(\cos ^2 x-2 \sin x \cos x\right) \mathrm{d} x=\mathrm{c} \)
\( \Rightarrow y+\log |y|+\mathrm{e}^x \cos ^2 x=\mathrm{c} \)
\( \text { At } x=0, y=1 \)
\( \therefore 1+\log 1+\mathrm{e}^0 \cos ^2 0=\mathrm{c} \)
\( \Rightarrow \mathrm{c}=2 \)
\( \therefore y+\log |y|+\mathrm{e}^x \cos ^2 x=2\)
\( \Rightarrow \frac{y+1}{y} \mathrm{~d} y+\mathrm{e}^x\left(\cos ^2 x-\sin 2 x\right) \mathrm{d} x=0\)
Integrating on both sides, we get
\( \int\left(1+\frac{1}{y}\right) \mathrm{d} y+\int \mathrm{e}^x\left(\cos ^2 x-2 \sin x \cos x\right) \mathrm{d} x=\mathrm{c} \)
\( \Rightarrow y+\log |y|+\mathrm{e}^x \cos ^2 x=\mathrm{c} \)
\( \text { At } x=0, y=1 \)
\( \therefore 1+\log 1+\mathrm{e}^0 \cos ^2 0=\mathrm{c} \)
\( \Rightarrow \mathrm{c}=2 \)
\( \therefore y+\log |y|+\mathrm{e}^x \cos ^2 x=2\)
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