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?

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

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?

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

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

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

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

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

Originally Posted by Prove It

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

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

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

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

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

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

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

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


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