{y}'- \sin{{2}}{x}+{y} \tan{{x}}={0}

Jerold 2021-08-31 Answered

ysin2x+ytanx=0

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Expert Answer

lamusesamuset
Answered 2021-09-01 Author has 93 answers

Solve the linear equation dy(x)dxsin(2x)+tan(x)y(x)=0:
Add sin(2x) to both sides:
dy(x)dx+tan(x)y(x)=sin(2x)
Let μ(x)=etan(x)dx=sec(x).
Multiply both sides by μ(x):
sec(x)dy(x)dx+(sec(x)tan(x))y(x)=sec(x)sin(2x)
INTERMEDIATE STEPS:
Take the integral:
tan(x)dx
Rewrite tan(x) as sin(x)cos(x):
sin(x)cos(x)
For the integrand sin(x)cos(x), substitute u=cos(x)anddu=sin(x)dx:
=1udu
Factor out constants:
=1udu
The integral of 1u is log(u):
=log(u)+constant
Substitute back for u=cos(x):
Answer:
=log(cos(x))+constant
Substitute tan(x)sec(x)=dsec(x)dx:
sec(x)dy(x)dx+dsec(x)dxy(x)=sec(x)sin(2x)
INTERMEDIATE STEPS:
Possible derivation:
ddx(sec(x))
Rewrite the expression:

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