What is a particular solution to the differential equation #dy/dx=cosxe^(y+sinx)# with #y(0)=0#?

Question

Aurora Alvarado · Accepted Answer

#y =ln (1/( 2 - e^(sin x) ))#

this is separable!!

#dy/dx=cosx \ e^(y+sinx)#

split the exponent...so #dy/dx=cosx \ e^y \ e^sinx#

#e^-y \ dy/dx=cosx \ e^sinx#

and this is separated, so next...integrate each side wrt x

#int \ e^-y \ dy/dx \ dx = int \ cosx \ e^sinx \ dx#

#int \ e^-y \ dy = int \ cosx \ e^sinx \ dx#

#- e^-y = color(red)( int \ cosx \ e^sinx \ dx) qquad triangle#

we can do the red RHS in a number of ways, the more laboured involving some kind of sub: #u = sin x#

but before we do that, just note that

#d/dx e^(f(x)) = f'(x) e^(f(x))# and therefore #int \ d/dx e^(f(x)) \ dx = int \ f'(x) e^(f(x)) \dx#

OR

# e^(f(x)) = int \ f'(x) e^(f(x)) \dx + C#

if you see that, brilliant. the job is done!

if not :-((

well, then we go with #triangle# but with a sub as mentioned. #u = sin x, du = cos x \ dx, color{blue}{u' = cos x}# so #triangle# becomes

#-e^-y = int \ u' \ e^u \ (1)/(u') \ du= int \ e^u \ du#

#implies -e^-y = e^u + C = e^(sin x) + C#

So, to tidy it all up....

#e^-y = C - e^(sin x) #

#ln( e^-y) =ln ( C - e^(sin x) )#

#-y =ln ( C - e^(sin x) )#

#y =ln (1/( C - e^(sin x) ))#

so #y_o = 0# means that

#0 =ln (1/( C - 1 ))# so #C = 2#

#y =ln (1/( 2 - e^(sin x) ))#

Emilia Aguilar · Answer

undefined The particular solution to the given differential equation $ \frac{dy}{dx} = \cos(x)e^{y+\sin(x)} $ with the initial condition $ y(0) = 0 $ is $ y = \ln(\cos(x)+1) - \sin(x) $.