Question

Aplique o teorema da valor inicial e determine a única função y=y (x), definida em R, tal que $\frac{dy}{dx} = xe^{4-x^2}$ no ponto (-2,1).

Answer 1

$\mathtt{ \frac{dy}{dx} = x {e}^{4 - {x}^{2} } } \\\mathtt{dy = x {e}^{4 - {x}^{2}}dx} \\\mathtt{\int dy= \int x{e}^{4 -{x}^{2}}dx}$

$\mathtt{\int x {e}^{4 -{x}^{2}} dx} = \mathtt{ - \dfrac{1}{2} \int - 2x {e}^{4 -{x}^{2}} dx} \\\mathtt{ = - \frac{1}{2}{e}^{4-{x}^{2}} + k}$

$\mathtt{\int dy= \int x{e}^{4 -{x}^{2}}dx} \\\mathtt{y(x) = -\frac{1}{2}{e}^{4 - {x}^{2} } + k }$

$\mathtt{y( - 2) =-\frac{1}{2} {e}^{4 - {1}^{2} } + k} \\\mathtt{1 = - \frac{1}{2} {e}^{3} + k} \\\mathtt{k = \frac{2+{e}^{3} }{2} }$

$\boxed{\boxed{\mathtt{y(x) = - \frac{1}{2}{e}^{4 - {x}^{2} }+\dfrac{2+{e}^{3}}{2}}}}$