Question

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}.

Forms of linear equations
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asked 2021-01-31
Linear equations of first order.
Solve the initial-value problem on the specified interval \({y}'+{y} \tan{{x}}= \sin{{2}}{x}{o}{n}{\left(-\frac{1}{{2}}\pi,\frac{1}{{2}}\pi\right)}\ \text{with}\ {y}={2}\ \text{when}\ {x}={0}\).

Answers (1)

2021-02-01

From the given linear differential equation, \(P(x)= \tan x\) and \(Q(x)=\sin2x.\)
Therefore, the integrating factor is given by
\({\left({x}\right)}={\int_{{0}}^{{x}}} \tan{{t}}{\left.{d}{t}\right.}={{\left[- \ln{ \cos{{t}}}\right]}_{{0}}^{{x}}}=- \ln{ \cos{{x}}}\).
So that, the solution is given by
\({y}={2} \cos{{x}}+ \cos{{x}}{\int_{{1}}^{{x}}}{\left( \cos{{t}}\right)}^{ -{{1}}} \sin{{2}}{t}{h}{t}\)
\(={2} \cos{{x}}+ \cos{{x}}{\int_{{1}}^{{x}}}{\left( \cos{{t}}\right)}^{ -{{1}}}{\left({2} \sin{{t}} \cos{{t}}\right)}{\left.{d}{t}\right.}\)
\(={2} \cos{{x}}+ \cos{{x}}{\int_{{1}}^{{x}}} \sin{{t}}{\left.{d}{t}\right.}\)
\(={2} \cos{{x}}+ \cos{{x}}{{\left[- \cos{{t}}\right]}_{{0}}^{{x}}}\)
\(={2} \cos{{x}}+ \cos{{x}}{\left({1}- \cos{{x}}\right)}\)
\(={4} \cos{{x}}-{2}{{\cos}^{2}{x}}\).
We can verify the function is the solution of the initial-value periblem since it satisfies the differential equation and the initial condition.

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