Let y = y(x) be the solution of the differential equation cosx[dy dx] + 2ysinx = sin2x, x …

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 :

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 = $$\sqrt 2  - 2$$

Solutions · 1