Question

Consider $g\left(x\right)=$ $\{\begin{array}{cc}sinx& 0\le x\le \frac{\pi }{2}\\ cosx& x>\frac{\pi }{2}\end{array}$ and a continuous function $y=f\left(x\right)$ satisfies $\frac{5dy}{dx}+5y=g\left(x\right)$ $f\left(0\right)=0$, then,

• A. $f\left(\frac{\pi }{4}\right)=\frac{e^{\frac{-\pi }{4}}}{10}$
• B. $f\left(\frac{\pi }{4}\right)=\frac{e^{\frac{-\pi }{4}}-1}{10}$
• C. $f\left(\frac{\pi }{4}\right)=\frac{e^{\frac{-\pi }{2}}+1}{10}$
• D. $f\left(\frac{\pi }{4}\right)=e^{\frac{-\pi }{2}}$

Answer 1

The correct option is B $f\left(\frac{\pi }{4}\right)=\frac{e^{\frac{-\pi }{4}}-1}{10}$

Given,

$\frac{5dy}{dx}+5y=g\left(x\right)$

$\frac{dy}{dx}+y=\frac{g\left(x\right)}{5}$

This is the linear diffrential equation

so integrated factor

I.F.=$e^{\int p.dx}$

i.f.=$e^{\int 1.dx}$

i.f.=$e^x$

So solution is

$y\times if=\int if\times qdx$

$y.e^x=\int \frac{1}{5}{e}^{x}Sinxdx$

$y.e^x=\frac{e^x}{10}(sinx-cosx)+c$$-\left(1\right)$

$0=\frac{-1}{10}+c$

$c=\frac{1}{10}$

$\left(1\right)\to y=\frac{1}{10}(sinx-cosx)+\frac{e^{-x}}{10}$

$y_{at\frac{\pi }{4}}=\frac{-1}{10}+\frac{e^{-\frac{\pi }{4}}}{10}$