Solve \frac{dy}{dx}=\cos(x+y)+\sin(x+y) My Attempt \frac{dy}{dx}=\sqrt{2}\cos(x+y-\frac{\pi}{4}) Set t=x+y-\frac{\pi}{4}\implies y=t-x+\frac{\pi}{4} \frac{dy}{dx}=\frac{dt}{dx}-1=\sqrt{2}\cos t \frac{dt}{dx}=\sqrt{2}\cos(t)+1 \int\frac{dt}{\sqrt{2}\cos(t)+1}=\int dx How do I proceed

ulcerat9jr

ulcerat9jr

Answered question

2022-04-21

Solve dydx=cos(x+y)+sin(x+y)
My Attempt
dydx=2cos(x+yπ4)
Set t=x+yπ4y=tx+π4
dydx=dtdx1=2cost
dtdx=2cos(t)+1
dt2cos(t)+1=dx
How do I proceed further to find the general solution ?

Answer & Explanation

Kathleen Keller

Kathleen Keller

Beginner2022-04-22Added 20 answers

Making x+y=udydx=dudx1
dudx=sinu+cosu+1
this is a separable DE
dusinu+cosu+1=dx
etc.
NOTE
sinu+cosu=2sin(u+π4)
now using the identities
cosu=2cos2(u2)1
sinu=2cos(u2)sin(u2)
we have
1cosu+sinu+1=12cos(u2)(cos(u2)+sin(u2))=12sin(u2)cos(u2)+sin(u2)+cos(u2)cos(u2)+sin(u2)

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