CoV14

Compute the first variation of $J(y)=\int_a^b (y'^2+2y)dx\,$

$\delta J(y, h)\,$	$= \frac{d}{d\varepsilon} J(y + \varepsilon h) \left.\right\|_{\varepsilon = 0}$
	$= \frac{d}{d\varepsilon} \int_a^b ((y^\prime + \varepsilon h^\prime)^2 + 2(y + \varepsilon h)) \ dx\left.\right\|_{\varepsilon = 0}$
	$= \frac{d}{d\varepsilon} \int_a^b (y^{\prime 2} + 2\varepsilon y^\prime h^\prime + \varepsilon^2 h^{\prime 2} + 2y + 2\varepsilon h) \ dx\left.\right\|_{\varepsilon = 0}$
	$= \int_a^b \frac{d}{d\varepsilon} (y^{\prime 2} + 2\varepsilon y^\prime h^\prime + \varepsilon^2 h^{\prime 2} + 2y + 2\varepsilon h) \ dx\left.\right\|_{\varepsilon = 0}$
	$= \int_a^b (2y^\prime h^\prime + 2\varepsilon h^{\prime 2} + 2h) \ dx\left.\right\|_{\varepsilon = 0}$
	$= \int_a^b (2y^\prime h^\prime + 2h) \ dx$

$\delta J(y, h)\,$	$= \frac{d}{d\varepsilon} J(y + \varepsilon h) \left.\right\|_{\varepsilon = 0}$
	$= \frac{d}{d\varepsilon} \int_a^b ((y^\prime + \varepsilon h^\prime)^2 + 2(y + \varepsilon h)) \ dx\left.\right\|_{\varepsilon = 0}$
	$= \frac{d}{d\varepsilon} \int_a^b (y^{\prime 2} + 2\varepsilon y^\prime h^\prime + \varepsilon^2 h^{\prime 2} + 2y + 2\varepsilon h) \ dx\left.\right\|_{\varepsilon = 0}$
	$= \int_a^b \frac{d}{d\varepsilon} (y^{\prime 2} + 2\varepsilon y^\prime h^\prime + \varepsilon^2 h^{\prime 2} + 2y + 2\varepsilon h) \ dx\left.\right\|_{\varepsilon = 0}$
	$= \int_a^b (2y^\prime h^\prime + 2\varepsilon h^{\prime 2} + 2h) \ dx\left.\right\|_{\varepsilon = 0}$
	$= \int_a^b (2y^\prime h^\prime + 2h) \ dx$