Question

.. y' = y tan 2x, y (0) = 2.​

Answer 1

Given that,

$\sf \: \dfrac{dy}{dx} = y \times tan(2x)$

When x = 0, y = 2.

The above differential equation can be solved by variable separable method.

$\implies \sf \: \dfrac{dy}{y} = tan(2x)dx \\ \\\implies \displaystyle \int \sf \: \dfrac{dy}{y} = \int tan(2x)dx \\ \\ \implies \displaystyle \sf log(y) = \dfrac{1}{2} \int tan(t)dt \\ \\ \implies \displaystyle \sf log(y) = \dfrac{1}{2} log(sec2x) + log(k) \\ \\ \implies \displaystyle \sf 2 log(y) = log(sec2x) + 2log(k) \\ \\ \implies \sf \: log(y {}^{2} ) = log(k {}^{2} sec2x) \\ \\ \implies \sf \: y {}^{2} cos2x = k {}^{2} - - - - - - - - (1)$

We know that, when x = 0, y = 2.

Thus,

$\implies \sf \: k {}^{2} = (2) {}^{2} cos(0) \\ \\ \implies \sf \: k = \pm 2$

Putting value of k in equation (1),

$\implies \sf \: y {}^{2} cos2x = 4 \\ \\ \implies \boxed{ \boxed{ \sf y \sqrt{cos2x} = \pm2 }}$

Thus, the particular solution of the above differential equation at y(0) = 2 is y√cos2x = ± 2.