Differential Equations – Variable Separable – JEE Main 22 Jan 2026 Shift 2

Solution:

Rearrange the differential equation to separate variables:
$$ \frac{\cos y}{1+2\sin y} dy = \frac{dx}{16(\sqrt{x+9\sqrt{x}})(4+\sqrt{9+\sqrt{x}})} $$

Integrate both sides:
**LHS Integration:**
Let $1 + 2\sin y = u \Rightarrow 2\cos y \, dy = du$.
$$ \int \frac{\cos y}{1+2\sin y} dy = \frac{1}{2} \int \frac{du}{u} = \frac{1}{2} \ln|1+2\sin y| $$

**RHS Integration:**
Let $4 + \sqrt{9+\sqrt{x}} = t$.
Differentiating with respect to $x$:
$$ dt = \frac{1}{2\sqrt{9+\sqrt{x}}} \cdot \frac{1}{2\sqrt{x}} dx $$
$$ dt = \frac{dx}{4\sqrt{x(9+\sqrt{x})}} = \frac{dx}{4\sqrt{9x+x\sqrt{x}}} = \frac{dx}{4\sqrt{9\sqrt{x}+x}} $$
From the question, the term in the denominator is $\sqrt{x+9\sqrt{x}}$, which matches the derivative term.
So, substitute $dx = 4\sqrt{x+9\sqrt{x}} \, dt$:
$$ \int \frac{4\sqrt{x+9\sqrt{x}} \, dt}{16(\sqrt{x+9\sqrt{x}})(t)} = \int \frac{4 dt}{16 t} = \frac{1}{4} \ln|t| $$
$$ = \frac{1}{4} \ln|4 + \sqrt{9+\sqrt{x}}| $$

**General Solution:**
$$ \frac{1}{2} \ln(1+2\sin y) = \frac{1}{4} \ln(4+\sqrt{9+\sqrt{x}}) + C $$
Multiply by 4 to simplify:
$$ 2 \ln(1+2\sin y) = \ln(4+\sqrt{9+\sqrt{x}}) + 4C $$

**Finding Constant C:**
Given $y(256) = \frac{\pi}{2}$.
Substitute $x = 256$ and $y = \frac{\pi}{2}$:
$$ \sqrt{x} = 16 \Rightarrow \sqrt{9+16} = 5 $$
$$ 2 \ln(1 + 2(1)) = \ln(4+5) + 4C $$
$$ 2 \ln 3 = \ln 9 + 4C $$
$$ \ln(3^2) = \ln 9 + 4C \Rightarrow \ln 9 = \ln 9 + 4C \Rightarrow C = 0 $$

**Finding $\alpha$:**
Given $y(49) = \alpha$.
Substitute $x = 49$:
$$ \sqrt{x} = 7 \Rightarrow \sqrt{9+7} = 4 $$
Equation becomes:
$$ 2 \ln(1+2\sin \alpha) = \ln(4+4) $$
$$ \ln(1+2\sin \alpha)^2 = \ln 8 $$
$$ (1+2\sin \alpha)^2 = 8 $$
Taking the positive root (since sine is bounded):
$$ 1+2\sin \alpha = \sqrt{8} = 2\sqrt{2} $$
$$ 2\sin \alpha = 2\sqrt{2} – 1 $$

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