Question

If \[ \left(\frac{2+\sin x}{1+y}\right)\frac{dy}{dx}=-\cos x \] and \[ y(0)=2, \] then find the value of \[ y\left(\frac{\pi}{2}\right). \]

• A. \(3\)
• B. \(4\)
• C. \(2\)
• D. \(1\)

Hint:

Whenever an integral contains a fraction of the form \[ \frac{f'(x)}{f(x)}, \] directly use: \[ \int \frac{f'(x)}{f(x)}dx=\ln|f(x)|+C \] This is one of the most frequently used formulas in differential equations.

Answer 1

The Correct Option is D

Concept: The given differential equation is a first-order separable differential equation. In such equations, all terms containing \(y\) are brought to one side and all terms containing \(x\) are brought to the other side. After separating variables, both sides are integrated independently.

Step 1: Rewriting the differential equation in separable form.
Given: \[ \left(\frac{2+\sin x}{1+y}\right)\frac{dy}{dx}=-\cos x \] Multiply both sides by \[ \frac{1+y}{2+\sin x} \] We get: \[ \frac{dy}{dx} = -\frac{(1+y)\cos x}{2+\sin x} \] Now divide both sides by \(1+y\): \[ \frac{1}{1+y}\frac{dy}{dx} = -\frac{\cos x}{2+\sin x} \] Multiplying throughout by \(dx\): \[ \frac{1}{1+y}\,dy = -\frac{\cos x}{2+\sin x}\,dx \] Thus, the variables are completely separated.

Step 2: Integrating both sides.
Integrating: \[ \int \frac{1}{1+y}\,dy = -\int \frac{\cos x}{2+\sin x}\,dx \] The left-hand side becomes: \[ \ln|1+y| \] For the RHS, let: \[ u=2+\sin x \] Then: \[ du=\cos x\,dx \] Therefore: \[ -\int \frac{\cos x}{2+\sin x}\,dx = -\int \frac{du}{u} \] Using the standard logarithmic integral: \[ \int \frac{du}{u}=\ln|u| \] Hence: \[ -\int \frac{du}{u} = -\ln|u| \] Substituting back: \[ -\ln|2+\sin x| \] Therefore, \[ \ln|1+y| = -\ln|2+\sin x|+C \]

Step 3: Simplifying the logarithmic expression.
Using the logarithmic identity: \[ \ln a+\ln b=\ln(ab) \] we obtain: \[ \ln|1+y|+\ln|2+\sin x|=C \] \[ \ln|(1+y)(2+\sin x)|=C \] Removing logarithm: \[ (1+y)(2+\sin x)=K \] where \(K\) is a constant.

Step 4: Using the initial condition \(y(0)=2\).
Substitute: \[ x=0,\qquad y=2 \] into the general solution: \[ (1+2)(2+\sin0)=K \] Since: \[ \sin0=0, \] we get: \[ 3\times2=K \] \[ K=6 \] Thus, the required particular solution becomes: \[ (1+y)(2+\sin x)=6 \]

Step 5: Finding \(y\left(\frac{\pi}{2}\right)\).
Substitute: \[ x=\frac{\pi}{2} \] Since: \[ \sin\frac{\pi}{2}=1, \] we get: \[ (1+y)(2+1)=6 \] \[ 3(1+y)=6 \] \[ 1+y=2 \] \[ y=1 \] Hence, \[ \boxed{ y\left(\frac{\pi}{2}\right)=1 } \]