Question

If $\frac{dy}{dx}+y\tan x=\sin 2x$ and $y\left(0\right)=1$, then $y\left(\pi \right)=?$

• A. $1$
• B. $-1$
• C. $-5$
• D. $5$

Answer 1

The correct option is C $-5$

$I.F.=e^{\int tanxdx}$

$⟹I.F.=e^{-lncosx}$

$I.F.=\frac{1}{cosx}$

Multiplying with $I.F.$ in the given equation,we get-

$\frac{dy}{cosxdx}+\frac{ysinx}{co{s}^{2}x}=2sinx$

$d\left(\frac{y}{cosx}\right)=2sinxdx$

Integrating both sides, we get-

$\frac{y}{cosx}=-2cosx+C$

$y\left(0\right)=1$

$1=-2+C$

$C=3$

For $x=\pi$ we get

$\frac{y}{-1}=2+3$

$y=-5$