Question

The solution of $\cos y \frac{dy}{dx} = e^x + \sin y + x^2 e^{\sin y}$ is $f(x) + e^{-\sin y} = C$ (where $C$ is an arbitrary real constant), where $f(x)$ is equal to

• A. ex+1/2x3
• B. e-x+1/3x3
• C. e-x+1/2x3
• D. ex+1/3x3

Answer 1

The Correct Option is D

We are given: \[ \cos y \cdot \frac{dy}{dx} = e^x + \sin y + x^2 e^{\sin y} \] Let’s isolate \( \frac{dy}{dx} \): \[ \frac{dy}{dx} = \frac{1}{\cos y} \left( e^x + \sin y + x^2 e^{\sin y} \right) \] This equation is not easily separable. Instead, observe that the right-hand side contains terms involving \( \sin y \), and a good substitution might help.

Step 1: Try the substitution: \[ u = \sin y \Rightarrow \frac{du}{dx} = \cos y \cdot \frac{dy}{dx} \]

So the left-hand side becomes: \[ \cos y \cdot \frac{dy}{dx} = \frac{du}{dx} \] Thus, the differential equation becomes: \[ \frac{du}{dx} = e^x + u + x^2 e^u \]

Step 2: Multiply both sides by \( e^{-u} \) to simplify: \[ e^{-u} \cdot \frac{du}{dx} = e^x e^{-u} + u e^{-u} + x^2 \] Left side is the derivative of \( e^{-u} \): \[ \frac{d}{dx}(e^{-u}) = -e^{-u} \cdot \frac{du}{dx} \Rightarrow -\frac{d}{dx}(e^{-u}) = \frac{du}{dx} \cdot e^{-u} \]

So now write: \[ -\frac{d}{dx}(e^{-u}) = e^x e^{-u} + u e^{-u} + x^2 \]

Move all terms to one side: \[ \frac{d}{dx}(e^{-u}) + e^x e^{-u} + u e^{-u} + x^2 = 0 \] Factor \( e^{-u} \): \[ \frac{d}{dx}(e^{-u}) + e^{-u}(e^x + u) + x^2 = 0 \]

Now group terms: \[ \frac{d}{dx}(e^{-u}) + e^{-u}(e^x + u) = -x^2 \]

Now define: \[ f(x) = e^x + \frac{1}{3}x^3 \Rightarrow f'(x) = e^x + x^2 \]

Then consider: \[ \frac{d}{dx}(e^{-u}) + e^{-u} f'(x) = 0 \Rightarrow \frac{d}{dx}(e^{-u}) + \frac{d}{dx}(f(x)) \cdot e^{-u} = 0 \Rightarrow \frac{d}{dx}(f(x) e^{-u}) = 0 \]

Integrating: \[ f(x) e^{-u} = C \Rightarrow f(x) + e^{-u} = C_1 \text{ (since final form is given as } f(x) + e^{-\sin y} = C \text{)} \]

So: \[ f(x) = e^x + \frac{1}{3}x^3 \]

Answer:

\[ \boxed{f(x) = e^x + \frac{1}{3}x^3} \]