Задача 72043 y`+ytgx=cos2x,y|x=n/4=1/2...

6488c5d9571d722cc71cc08c

13.06.2023 22:44:42

y`+ytgx=cos2x,y|x=n/4=1/2

5f3ea7e3faf909182968ddd9

14.06.2023 08:18:33

★

y`+ytgx=cos2x
y=u*v

u`*v+u*v`+u*vtgx=cos2x

u`*v+u*(v`+vtgx)=cos2x

(v`+vtgx)=0
⇒
u`*v=cos2x


1)v`+vtgx=0
dv/v=-tgxdx
∫ dv/v=- ∫ tgxdx
ln|v|=ln|cosx|
v=cosx


2)
u`*cosx=cos2x

du=(cos2x/cosx)dx

u= ∫ (cos2x/cosx)dx= ∫ (2cos^2x-1)/cosxdx= ∫ (2cosx-(1/cosx))dx

u=2sinx-ln|tg(π/4)+x/2)|+C

y=(2sinx-ln|tg(π/4)+x/2)|+C)*cosx


x=π/4

y=1/2

1/2=(2sin(π/4)-ln|tg(π/4)+(π/4)/2)|+C)*cos(π/4)

C=