 # 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 .

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From the given linear differential equation, $P\left(x\right)=\mathrm{tan}x$ and $Q\left(x\right)=\mathrm{sin}2x.$
Therefore, the integrating factor is given by
$\left(x\right)={\int }_{0}^{x}\mathrm{tan}tdt={\left[-\mathrm{ln}\mathrm{cos}t\right]}_{0}^{x}=-\mathrm{ln}\mathrm{cos}x$.
So that, the solution is given by
$y=2\mathrm{cos}x+\mathrm{cos}x{\int }_{1}^{x}{\left(\mathrm{cos}t\right)}^{-1}\mathrm{sin}2tht$
$=2\mathrm{cos}x+\mathrm{cos}x{\int }_{1}^{x}{\left(\mathrm{cos}t\right)}^{-1}\left(2\mathrm{sin}t\mathrm{cos}t\right)dt$
$=2\mathrm{cos}x+\mathrm{cos}x{\int }_{1}^{x}\mathrm{sin}tdt$
$=2\mathrm{cos}x+\mathrm{cos}x{\left[-\mathrm{cos}t\right]}_{0}^{x}$
$=2\mathrm{cos}x+\mathrm{cos}x\left(1-\mathrm{cos}x\right)$
$=4\mathrm{cos}x-2{\mathrm{cos}}^{2}x$.
We can verify the function is the solution of the initial-value periblem since it satisfies the differential equation and the initial condition.