Let y = y(x) be the solution of the differential equation cosx${{dy} \over {dx}}$ + 2ysinx = sin2x, x $ \in $ $\left( {0,{\pi \over 2}} \right)$. If y$\left( {{\pi \over 3}} \right)$ = 0, then y$\left( {{\pi \over 4}} \right)$ is equal to :

Question

Let y = y(x) be the solution of the differential equation

cosx${{dy} \over {dx}}$ + 2ysinx = sin2x, x $ \in $ $\left( {0,{\pi \over 2}} \right)$.

If y$\left( {{\pi \over 3}} \right)$ = 0, then y$\left( {{\pi \over 4}} \right)$ is equal to :

${1 \over {\sqrt 2 }} - 1$

${\sqrt 2 - 2}$

${2 - \sqrt 2 }$

${2 + \sqrt 2 }$

Solution

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