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

We have a first-order differential equation with an initial condition — this is an Initial Value Problem. The equation is separable.

Step 1: Separate variables

Step 2: Integrate both sides

Step 3: Apply initial condition $y(0)=1$

Thus: …

The problem gives us a differential equation and an initial condition — that's an Initial Value Problem (IVP). The key idea: we can find a specific solution curve that passes through the point $(0,1)$, and then read off its height at $x = \pi/2$.

The equation is:

Notice that the variables $x$ and $y$ are not mixed together in a complicated way. The factor $\frac{2+\sin x}{1+y}$ is a product of a function of $x$ alone and a function of $y$ alone. This is the hallmark of a separable differential equation — we can rearrange it so that all $y$ terms are on one side and all $x$ terms on the other, then integrate.

Let's work through it step by step.

1. Separate the variables. Multiply both sides by $(1+y)$ and divide by $(2+\sin x)$ to isolate $dy/dx$:

Now divide both sides by $(1+y)$ (assuming $1+y \neq 0$ — we'll check later) and multiply by $dx$:

The variables are now separated: left side depends only on $y$, right side only on $x$.

2. Integrate both sides.

The left integral is straightforward:

For the right integral, notice that the numerator $\cos x$ is exactly the derivative of $\sin x$, and the denominator is $2+\sin x$. This suggests a simple substitution: let $u = 2+\sin x$, then $du = \cos x\,dx$. So:

Therefore, the integrated equation becomes:

where $C = C_2 - C_1$ is a combined constant.

3. Simplify using logarithm properties. Bring the negative inside as a power:

Exponentiate both sides (remember $e^{\log A} = A$):

Let $K = e^C > 0$, so:

Since $2+\sin x$ is always positive (minimum value is $2-1=1$), we can drop the absolute value on the denominator. The absolute value on $1+y$ can be handled by allowing $K$ to be any nonzero constant (positive or negative), because $1+y$ could be positive or negative. So we write:

$1+y = \frac{A}{2+\sin x}$ …

Method: Separable initial-value problem evaluated at a point

Use this when a separable DE has an initial condition and you must return a specific output value.

Steps

Step 1: Separate and integrate

Split the variables and integrate. Watch for a right side whose numerator is the derivative of its denominator — then $\displaystyle\int\frac{f'(x)}{f(x)}\,dx=\log|f(x)|$.

Step 2: Fix the constant with the initial condition …

Common Mistakes

Mistake 1: Missing that the $x$-integral is a logarithm

Why it's wrong: $\int\frac{\cos x}{2+\sin x}\,dx=\log|2+\sin x|$ because the numerator is the derivative of the denominator. Correct approach: use the $\frac{f'}{f}$ pattern.

Mistake 2: Combining constants sloppily

Why it's wrong: applying $y(0)=1$ gives $C=\log 4$; a careless constant makes the final value wrong. Correct approach: solve for $C$ exactly, then evaluate. …