First order DE

**Jones** · February 12th 2010, 05:26 AM

Hi,

I have the following DE:

$y' + ycos~x = sin2x$

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

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

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

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

**Sudharaka** · February 12th 2010, 05:50 AM

Originally Posted by Jones

Hi,

I have the following DE:

$y' + ycos~x = sin2x$

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

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

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

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

Dear Jones,

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

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

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

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

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

$=2\int{lnu~du}$

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

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

Hope this will help you.

**Jones** · February 12th 2010, 06:17 AM

Thank you very much Sudharaka.

I hate myself for not seeing that obvious substitution

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