Question

For the given differential equation, find the particular solution satisfying the given condition: $\frac{dy}{dx} +2y\ tan\ x = sin\ x; y = 0\ when\ x = \frac{\pi}{3}$

Answer 1

To find the particular Solution for the given differential equation, satisfying the given condition: $\frac{dy}{dx} +2y\ tan\ x = sin\ x; y = 0\ when\\\ x = \frac{\pi}{3}$

The equation is in the form of linear differential equation.
$\frac{dy}{dx} + Py = Q$
Solution:

$y.I.F. = \int(Q \times I.F.)dx + C$

To find I.F.:

${e}^{\int Pdx} \\ \\ = {e}^{\int2 \: tan \: x \: dx} \\ \\ = {e}^{- 2log \: cos\: x} \\ \\ = {(cos \: x)}^{ - 2} \\ \\ = \frac{1}{ {cos}^{2}x }$
Solution

$y\times I.F. = \int(Q \times I.F.)dx + C \\ \\ y. \frac{1}{{cos}^{2} x } = \int \: sin \: x \times \frac{1}{{cos}^{2} x } \: dx \\ \\ \frac{y}{ {cos}^{2}x } =\int \: sec \: x \: tan \: x \: dx + c \\ \\ \\\frac{y}{ {cos}^{2}x } = sec \: x + C \\ \\ y = cos \: x + C\: {cos}^{2} x$

for particular Solution:

put the value of x =π/3,y=0

$0 = cos \: \frac{\pi}{3} + C\: {cos}^{2} \frac{\pi}{3} \\ \\0=\frac{1}{2}+\frac{C}{4}\\\\C=-2$

Put the value of C,in

$y = cos \: x + C\: {cos}^{2} x \\ \\y = cos \: x -2\: {cos}^{2} x$