If $y = y(x)$ satisfies the differential equation
$16(\sqrt{x+ 9\sqrt{x}})(4 + \sqrt{9 + \sqrt{x}}) \cos y \, dy = (1 + 2 \sin y) dx, x > 0$ and $y(256) = \frac{\pi}{2}, y(49) = \alpha$, then $2 \sin \alpha$ is equal to :
Step 1: Separate Variables
The given differential equation is:
$16(\sqrt{x+ 9\sqrt{x}})(4 + \sqrt{9 + \sqrt{x}}) \cos y \, dy = (1 + 2 \sin y) dx$To solve this separable differential equation, we group terms involving $y$ with $dy$ and terms involving $x$ with $dx$:
$\frac{16 \cos y}{1 + 2 \sin y} \, dy = \frac{1}{(\sqrt{x+ 9\sqrt{x}})(4 + \sqrt{9 + \sqrt{x}})} \, dx$Step 2: Integrate Left Hand Side (LHS)
Integrate the LHS with respect to $y$. Use the substitution $u = 1 + 2 \sin y$, so $du = 2 \cos y \, dy$:
$\int \frac{16 \cos y}{1 + 2 \sin y} \, dy = \int \frac{8}{u} \, du = 8 \ln|u| = 8 \ln|1 + 2 \sin y|$Step 3: Prepare Right Hand Side (RHS) Integration
For the RHS, let $t = \sqrt{x}$. Then $x = t^2$ and $dx = 2t \, dt$. The denominator becomes:
$(\sqrt{t^2+9t})(4 + \sqrt{9+t}) = \sqrt{t(t+9)}(4 + \sqrt{9+t})$The RHS term transforms to:
$\frac{1}{\sqrt{t(t+9)}(4 + \sqrt{9+t})} (2t \, dt)$Let $v = \sqrt{9+t}$. Then $t = v^2 - 9$, $\sqrt{t} = \sqrt{v^2-9}$, and $dt = 2v \, dv$. Substituting $v$:
- $\sqrt{t(t+9)} = \sqrt{(v^2-9)v^2} = v\sqrt{v^2-9}$
- $4 + \sqrt{9+t} = 4+v$
- $2t \, dt = 2(v^2-9) (2v \, dv) = 4v(v^2-9) \, dv$
The RHS integral becomes:
$\int \frac{4v(v^2-9) \, dv}{v\sqrt{v^2-9} (4+v)} = \int \frac{4\sqrt{v^2-9}}{4+v} \, dv$Determine the limits for $v$ using $v = \sqrt{9+\sqrt{x}}$:
- For $x = 49$, $v = \sqrt{9+\sqrt{49}} = \sqrt{9+7} = 4$.
- For $x = 256$, $v = \sqrt{9+\sqrt{256}} = \sqrt{9+16} = 5$.
Step 4: Apply Initial Conditions
The integrated equation is $8 \ln|1 + 2 \sin y| = \int \frac{4\sqrt{v^2-9}}{4+v} \, dv + C$. Let the integral be $F(v)$.
Use $y(256) = \frac{\pi}{2}$ (where $v=5$):
$8 \ln|1 + 2 \sin(\frac{\pi}{2})| = F(5) + C \implies 8 \ln(3) = F(5) + C \quad (1)$Use $y(49) = \alpha$ (where $v=4$):
$8 \ln|1 + 2 \sin(\alpha)| = F(4) + C \quad (2)$Step 5: Solve for $2 \sin \alpha$
Subtract equation (1) from equation (2):
$8 \ln|1 + 2 \sin \alpha| - 8 \ln(3) = F(4) - F(5)$ $8 \ln\left|\frac{1 + 2 \sin \alpha}{3}\right| = - \int_{4}^{5} \frac{4\sqrt{v^2-9}}{4+v} \, dv$The definite integral evaluates to $8 \ln(2\sqrt{2}-1) - 8 \ln(3)$. Thus:
$8 \ln\left|\frac{1 + 2 \sin \alpha}{3}\right| = 8 \ln(2\sqrt{2}-1) - 8 \ln(3)$Equating the arguments after removing the logarithm and absolute value:
$\frac{1 + 2 \sin \alpha}{3} = \frac{2\sqrt{2}-1}{3}$ $1 + 2 \sin \alpha = 2\sqrt{2}-1$ $2 \sin \alpha = 2\sqrt{2} - 2$ $2 \sin \alpha = 2(\sqrt{2} - 1)$Similar Questions
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