Question

If $e^{\sin x}$ is an integrating factor of \[ \frac{dy}{dx}+(y-1)\cos x=e^{-\sin x}\cos x, \] then the general solution is:

• A. $(e^{\sin x}+c)e^{-\sin x}$
• B. $(\sin x+e^{\sin x}+c)e^{-\sin x}$
• C. $(\cos x+e^{\sin x}+c)e^{-\sin x}$
• D. $(\cos x+e^{\cos x}+c)e^{-\sin x}$

Hint:

After multiplying by IF, left side becomes exact derivative.

Answer 1

The Correct Option is B

Concept: A linear differential equation: \[ \frac{dy}{dx}+Py=Q \] has integrating factor: \[ IF=e^{\int Pdx} \]

Step 1: Rewrite equation.
\[ \frac{dy}{dx}+y\cos x=\cos x+e^{-\sin x}\cos x \] So, \[ P=\cos x \] Given IF: \[ IF=e^{\sin x} \]

Step 2: Multiply by IF.
\[ e^{\sin x}\frac{dy}{dx}+y e^{\sin x}\cos x = e^{\sin x}\cos x + \cos x \] Left side becomes: \[ \frac{d}{dx}(y e^{\sin x}) \]

Step 3: Integrate both sides.
\[ \frac{d}{dx}(y e^{\sin x}) = e^{\sin x}\cos x + \cos x \] \[ y e^{\sin x}=\int e^{\sin x}\cos x\,dx+\int \cos x\,dx \] \[ = e^{\sin x}+\sin x + C \]

Step 4: Solve for $y$.
\[ y=(\sin x+e^{\sin x}+C)e^{-\sin x} \]