Question

Please show me how to solve the initial value problem y'=tanx y(pi/4)=3

Answer 1

The ODE is separable, i.e. you can write

$\dfrac{\mathrm dy}{\mathrm dx}=\tan x\iff\mathrm dy=\tan x\,\mathrm dx$

Integrating both sides gives the general solution.

$\displaystyle\int\mathrm dy=\int\tan x\,\mathrm dx$
$y=-\ln|\cos x|+C$

Given that $y\left(\dfrac\pi4\right)=3$, we have

$3=-\ln\left|\cos\dfrac\pi4\right|+C$
$3=-\ln\dfrac1{\sqrt2}+C$
$3-\ln\sqrt2=C$


and so the particular solution to the IVP is

$y=-\ln|\cos x|+3-\ln\sqrt2$
$y=3-\ln|\sqrt2\cos x|$