"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

Accepted Answer

"cosx${{dy} \over {dx}}$ + 2ysinx = sin2x

$\Rightarrow$ ${{dy} \over {dx}} + 2y\tan x = 2\sin x$

I.F = ${e^{\int {2\tan xdx} }}$ = sec² x

y.sec² x = ${\int {2\sin x{{\sec }^2}xdx} }$

$\Rightarrow$ ysec²x = ${\int {2\tan x\sec xdx} }$

$\Rightarrow$ ysec²x = 2secx + c

Given at x = ${\pi \over 3}$, y = 0

$\Rightarrow$ 0 = $2\sec {\pi \over 3} + c$

$\Rightarrow$ c = -4

ysec²x = 2secx - 4

Here put x = ${\pi \over 4}$

y.2 = $2\sqrt 2$ - 4

$\Rightarrow$ y = $\sqrt 2 - 2$"

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 :

Solution

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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 :

Solution

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More Differential Equations questions

Drill 25 more like these. Every day. Free.

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 :