y'=\cos^{2}x \cos y

Jessie Lee 2021-12-27 Answered
y=cos2xcosy
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Expert Answer

porschomcl
Answered 2021-12-28 Author has 28 answers
Given, y=cos2xcosy
dydx=cos2xcosy
dycosy=cos2xdx
Integrate both side, we get
dycosy=cos2x dx
secydy=(1+cos2x2)dx
ln(secx+tany)=12(x+sin2x2)+c
ln(secy+tanx)=x2+sin2x4+c
Hence, solution of above differential equation is:
ln(secy+tany)=x2+sin2x4+c
secy dy=secy(secy+tany)(secy+tany)dy=(sec2y+secytany)secy+tanydy
Let, v=secy+tany
differentiate both side, we get
dv=(secytany+sec2y)dy
1vdv=lnv+c
=ln(secy+tany)+c
secydy=ln(secy+tany)+c

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Paineow
Answered 2021-12-29 Author has 30 answers
dydx=cos2xcosy
Separate the variable x and y both side.
dycosy=cos2x dx
Simplify secy dy=cos2xdx
Integrate both side of equation
secy dy=cos2x dx
ln|tany+secy|=(1+cos2x)2dx
=12(1+cos2x)dx
ln|tany+secy|=12[x+sin2x2]+c

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karton
Answered 2022-01-09 Author has 439 answers

dydx=cos2xcosydycosy=cos2xdxdycosy=cos2x dxsecydy=(1+cos2x2)dxln(secx+tany)=12(x+sin2x2)+cln(secy+tanx)=x2+sin2x4+cln(secy+tany)=x2+sin2x4+csecy dy=secy(secy+tany)(secy+tany)dy=(sec2y+secytany)secy+tanydyLet, v=secy+tanydv=(secytany+sec2y)dy1vdv=lnv+c=ln(secy+tany)+csecydy=ln(secy+tany)+c

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