Question

Let y = y(x) be the solution curve of the differential equation
$\sin(2x^2) \log_e(\tan(x^2)) \,dy + (4xy - 4\sqrt{2}x\sin(x^2 - \frac{\pi}{4})) \,dx = 0, \quad 0 < x < \sqrt{\frac{\pi}{2}}$
which passes through the point $(\sqrt{\frac{π}{6}},1)$. Then $|y(\sqrt{\frac{π}{3}})|$
is equal to _______.

Answer 1

The correct answer is 1
$\frac{dy}{dx} + y\left(\frac{4x}{\sin(2x^2) \ln(\tan(x^2))}\right) = \frac{4\sqrt{2}x\sin\left(x^2 - \frac{\pi}{4}\right)}{\sin(2x^2) \ln(\tan(x^2))}$
$I.F. = e^{\int\frac{4x}{\sin(2x^2) \ln(\tan(x^2))} \,dx}$
$= e^{In|In(\tan x^2)} = In(\tan x^2)$
∴$$\int d(y \ln(\tan(x^2))) = \int \frac{4\sqrt{2}x\sin\left(x^2 - \frac{\pi}{4}\right)}{\sin(2x^2)} \,dx
⇒ $y \ln(\tan(x^2)) = \ln\left|\frac{\sec^2(x) + \tan(x)}{\csc^2(x) - \cot(x)}\right| + C$
$In (\frac{1}{\sqrt3}) = In(\frac{\frac{3}{\sqrt3}}{2-\sqrt3})+C$
$e = \ln\left(\frac{1}{\sqrt{3}}\right) - \ln\left(\frac{\sqrt3}{2-\sqrt3}\right)$
For $y(\sqrt{\frac{π}{3}})$
$y \ln(\sqrt{3}) = \ln\left|\frac{2 + \frac{\sqrt{3}}{2}}{\frac{\sqrt{3}}{2} + \frac{1}{\sqrt{3}}}\right| + \ln\left(\frac{1}{\sqrt{3}}\right) - \ln\left(\frac{\sqrt{3}}{2\sqrt{3}}\right)$
= $\ln(2 + \sqrt{3}) + \ln\left(\frac{1}{\sqrt{3}}\right) + \ln\left(\frac{1}{\sqrt{3}}\right) - \ln\left(\frac{\sqrt3}{2-\sqrt3}\right)$
$⇒ y\ In \sqrt3 = \ In (\frac{1}{\sqrt3})$
$⇒ \frac{y}{2}\ In 3 = -\frac{1}{2} \ In\ 3$
⇒ y = 1
$∴ |y(\sqrt{\frac{π}{3}})|$
= 1