Question

The solution of the differential equation \(e^{-x}(y+1)\text{d}y+(cos^2x-sin2x)y\,\text{d}x = 0\), given that \(y = 1\) when \(x = 0\) is

• A. \(logy+\frac{1}{y}+e^xcos^2x = 1\)
• B. \(logy+y+e^xcos^2x = 2\)
• C. \((y+1)+e^xcos^2x = 2\)
• D. \(log(y+\frac{1}{y})+e^xcos^2x = 1\)

Hint:

Separate variables and spot that e^x cos^2 x differentiates to e^x(cos^2 x - sin 2x).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
The equation can be written in separated form, with all y terms on one side and all x terms on the other.

Step 2: Separate the variables
\[ e^{-x}(y + 1)\,dy = -(\cos^2 x - \sin 2x)\,y\,dx \Rightarrow \frac{y + 1}{y}\,dy = -e^{x}(\cos^2 x - \sin 2x)\,dx \]

Step 3: Integrate
Left side: \(\int \left(1 + \frac1y\right)dy = y + \log y\).
For the right side note that \(\dfrac{d}{dx}\left(e^x\cos^2 x\right) = e^x\cos^2 x - e^x\cdot 2\sin x\cos x = e^x(\cos^2 x - \sin 2x)\).
So the right side integrates to \(-e^x\cos^2 x\), and
\[ y + \log y + e^x\cos^2 x = C \]

Step 4: Use the condition
At \(x = 0, y = 1\): \(1 + 0 + 1 = C\), so \(C = 2\).
\[ \log y + y + e^x\cos^2 x = 2 \]
This is option (B). Option (A) has \(1/y\) instead of y and C = 1, which does not satisfy the given point.

Final Answer:
The solution is \(\log y + y + e^x\cos^2 x = 2\), option (B). \[ \boxed{\log y + y + e^x\cos^2 x = 2} \]