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Let a solution y=y(x) of the differentia...

Let a solution `y=y(x)` of the differential equation `(dy)/(dx)cosx+y sin x=1` satisfy y(0)=1
Statement-1: `y(x)=sin((pi)/(4)+x)`
Statement-2: The integrating factor of the given differential equation is sec x.

A

Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-1.

B

Statement-1 is True, Statement-2 is True, Statement-2 is not a correct explanation for Statement-1.

C

Statement-1 is True, Statement-2 is False.

D

Statement-1 is False, Statement-2 is True.

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To solve the differential equation \(\frac{dy}{dx} \cos x + y \sin x = 1\) with the initial condition \(y(0) = 1\), we will follow these steps: ### Step 1: Rewrite the equation We start with the given differential equation: \[ \frac{dy}{dx} \cos x + y \sin x = 1 \] We can divide the entire equation by \(\cos x\) to simplify it: \[ \frac{dy}{dx} + y \frac{\sin x}{\cos x} = \frac{1}{\cos x} \] This simplifies to: \[ \frac{dy}{dx} + y \tan x = \sec x \] ### Step 2: Identify \(p\) and \(q\) This is now in the standard linear form: \[ \frac{dy}{dx} + p(x)y = q(x) \] where \(p(x) = \tan x\) and \(q(x) = \sec x\). ### Step 3: Find the integrating factor The integrating factor \(\mu(x)\) is given by: \[ \mu(x) = e^{\int p(x) \, dx} = e^{\int \tan x \, dx} \] The integral of \(\tan x\) is: \[ \int \tan x \, dx = -\ln(\cos x) = \ln(\sec x) \] Thus, the integrating factor is: \[ \mu(x) = e^{\ln(\sec x)} = \sec x \] ### Step 4: Multiply the equation by the integrating factor Now we multiply the entire differential equation by \(\sec x\): \[ \sec x \frac{dy}{dx} + y \sec x \tan x = 1 \] This can be rewritten as: \[ \frac{d}{dx}(y \sec x) = 1 \] ### Step 5: Integrate both sides Integrating both sides with respect to \(x\): \[ y \sec x = \int 1 \, dx = x + C \] Thus, we have: \[ y \sec x = x + C \] or \[ y = (x + C) \cos x \] ### Step 6: Apply the initial condition We use the initial condition \(y(0) = 1\): \[ 1 = (0 + C) \cos(0) \] Since \(\cos(0) = 1\), we get: \[ 1 = C \] Thus, \(C = 1\). ### Step 7: Write the final solution Substituting \(C\) back into the equation: \[ y = (x + 1) \cos x \] ### Step 8: Verify the statements 1. **Statement 1**: \(y(x) = \sin\left(\frac{\pi}{4} + x\right)\) is incorrect. The correct form we derived is \(y = (x + 1) \cos x\). 2. **Statement 2**: The integrating factor of the given differential equation is indeed \(\sec x\), which is correct. ### Conclusion - **Statement 1** is false. - **Statement 2** is true.

To solve the differential equation \(\frac{dy}{dx} \cos x + y \sin x = 1\) with the initial condition \(y(0) = 1\), we will follow these steps: ### Step 1: Rewrite the equation We start with the given differential equation: \[ \frac{dy}{dx} \cos x + y \sin x = 1 \] We can divide the entire equation by \(\cos x\) to simplify it: ...
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