Linear equations of first order. Solve the initial-value problem on the specified interval {y}'+{y} tan{{x}}= sin{{2}}{x}{o}{n}{left(-frac{1}{{2}}pi,frac{1}{{2}}piright)} text{with} {y}={2} text{when} {x}={0}.

Rui Baldwin 2021-01-31 Answered

Linear equations of first order.
Solve the initial-value problem on the specified interval y+ytanx=sin2x on(12π,12π) with y=2 when x=0.

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Expert Answer

bahaistag
Answered 2021-02-01 Author has 101 answers

From the given linear differential equation, P(x)=tanx and Q(x)=sin2x.
Therefore, the integrating factor is given by
(x)=0xtantdt=[lncost]0x=lncosx.
So that, the solution is given by
y=2cosx+cosx1x(cost)1sin2tht
=2cosx+cosx1x(cost)1(2sintcost)dt
=2cosx+cosx1xsintdt
=2cosx+cosx[cost]0x
=2cosx+cosx(1cosx)
=4cosx2cos2x.
We can verify the function is the solution of the initial-value periblem since it satisfies the differential equation and the initial condition.

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