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Consider the differential equation

$W^{\prime }\left( s\right) -2iH\left( s\right) W\left( s\right) -1=0$, $% s\in I\subset \mathbb{R} $

and consider the functions

$F\left( s\right) =\int\limits_{0}^{s}\sin \left( 2\int\limits_{0}^{u}H\left( t\right) dt\right) du$

$G\left( s\right) =\int\limits_{0}^{s}\cos \left( 2\int\limits_{0}^{u}H\left( t\right) dt\right) du$

where $H:I\rightarrow %TCIMACRO{\U{211d} }% %BeginExpansion \mathbb{R} %EndExpansion $ is functions real of one variable

Show that the general solution to the above differential equation is given by

$W\left( s\right) =\left\{ \left( F\left( s\right) -c_{1}\right) +i\left( G\left( s\right) +c_{2}\right) \right\} \left( F^{\prime }\left( s\right) -iG^{\prime }\left( s\right) \right) $

Can anyone help me solve this differential equation? Well I just know the ode's that I know do not have complex coefficients.

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You treat the equation as if it were real. Multiply by $\exp\bigl(-2\,i\int_0^sH(t)\,dt\bigr)$ to get $$ \Bigl(\exp\bigl(-2\,i\int_0^sH(t)\,dt\bigr)\,W\Bigr)'=\exp\bigl(-2\,i\int_0^sH(t)\,dt\bigr). $$ Integrate and remember that $$ \exp\Bigl(-2\,i\int_0^sH(t)\,dt\Bigr)=\cos\Bigl(2\int_0^sH(t)\,dt\Bigr)-\sin\Bigl(2\int_0^sH(t)\,dt\Bigr)i. $$

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  • $\begingroup$ I may be mistaken, but this derivative is strange. Why the derivative of $\int_0^sH(t)dt$ with respect to $s$ is not necessarily $H$. Unless it is in relation to $t$, but if it is in relation to $t$ we do not get the equality above $\endgroup$ – Mancala Oct 29 '18 at 19:41
  • $\begingroup$ The fundamental Theorem of Calculus says that if $f$ is continuous, then$$\frac{d}{dx}\int_a^xf(t)\,dt=f(x).$$ $\endgroup$ – Julián Aguirre Oct 29 '18 at 22:18

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