solve the differential equation by separation of variables dy/dx = (cos x)e^(y+sinx)

**-DQ-** · Mar 29th 2010, 03:54 PM

dy/dx = (cos x)e^(y+sinx)

i mean i put the y's and x's on the same side but couldn't proceed from it. guidance?

**Krizalid** · Mar 29th 2010, 04:14 PM

$e^{y+\sin x}=e^{y}e^{\sin x}.$

**Prove It** · Mar 29th 2010, 05:32 PM

Originally Posted by -DQ-

dy/dx = (cos x)e^(y+sinx)

i mean i put the y's and x's on the same side but couldn't proceed from it. guidance?

$\frac{dy}{dx} = \cos{(x)}e^{y + \sin{(x)}}$

$\frac{dy}{dx} = \cos{(x)}e^{\sin{(x)}}e^{y}$

$e^{-y}\,\frac{dy}{dx} = \cos{(x)}e^{\sin{(x)}}$

$\int{e^{-y}\,\frac{dy}{dx}\,dx} = \int{\cos{(x)}e^{\sin{(x)}}\,dx}$

$\int{e^{-y}\,dy} = \int{e^u\,du}$ upon making the substitution $u = \sin{(x)}$ .

**-DQ-** · Mar 29th 2010, 05:49 PM

Originally Posted by Prove It

$\frac{dy}{dx} = \cos{(x)}e^{y + \sin{(x)}}$

$\frac{dy}{dx} = \cos{(x)}e^{\sin{(x)}}e^{y}$

$e^{-y}\,\frac{dy}{dx} = \cos{(x)}e^{\sin{(x)}}$

$\int{e^{-y}\,\frac{dy}{dx}\,dx} = \int{\cos{(x)}e^{\sin{(x)}}\,dx}$

$\int{e^{-y}\,dy} = \int{e^u\,du}$ upon making the substitution $u = \sin{(x)}$ .

I did this then i did the integral. i got

ln(e^y) + C = e^(sinx) + C

y = e^(sinx) + C [not same C but its still a constant)

however my answer sheet says im wrong

**Prove It** · Mar 29th 2010, 05:55 PM

Originally Posted by -DQ-

I did this then i did the integral. i got

ln(e^y) + C = e^(sinx) + C

y = e^(sinx) + C [not same C but its still a constant)

however my answer sheet says im wrong

$\int{e^{-y}\,dy} \neq \ln{e^y}$ .

Think about $\int{e^{ax}\,dx}$ . What does that equal? What is $a$ in this case?

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