# Thread: First order DE

1. ## First order DE

Hi,

I have the following DE:

$\displaystyle y' + ycos~x = sin2x$

So i have $\displaystyle e^{\int cos~x} = e^{sinx}$

Multiply both sides by this factor and integrate. Gives me.

$\displaystyle ye^{sinx} = \int e^{sinx} \cdot 2sin~x\cdot cos~x$

This is where im stuck, how to i integrate the right hand side?

2. Originally Posted by Jones
Hi,

I have the following DE:

$\displaystyle y' + ycos~x = sin2x$

So i have $\displaystyle e^{\int cos~x} = e^sinx$

Multiply both sides by this factor and integrate. Gives me.

$\displaystyle ye^{sinx} = \int e^{sinx} \cdot 2sin~x\cdot cos~x$

This is where im stuck, how to i integrate the right hand side?
Dear Jones,

$\displaystyle \int{e^{sinx}sin2x~dx}$

$\displaystyle Substitute,~u=e^{sinx}\Rightarrow{sinx=lnu}$

$\displaystyle du=e^{sinx}cosx~dx$

$\displaystyle =\int{u\times{2}\times{sinx}\times{cosx}\frac{du}{ ucosx}}$

$\displaystyle =\int{2\ln{u}~du}$

$\displaystyle =2\int{lnu~du}$

$\displaystyle =2\left[ulnu-u\right]+C~;~Where~C~is~an~arbitary~constant.$

$\displaystyle =2\left[e^{sinx}sinx-e^{sinx}\right]+C$