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

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.