Question

Determine a solução geral da equação diferencial ordinária linear de 1ª ordem y' - 2y = e^2x

Answer 1

$Sabemos\ que: \\\\ \frac{d}{dx}y-2y=e^{2x} \ --\ \textgreater \ \ p(x)=-2 \ \| \ q(x)=e^{2x} \\\\ Calculando\ o\ fator\ integrante: \\\\ \mu'(x)=\mu(x).p(x) \\\\ \frac{d}{dx}\mu(x)=\mu(x).(-2) \ --\ \textgreater \ \ \frac{\frac{d}{dx}\mu(x)}{\mu(x)}= \frac{\mu(x).(-2)}{\mu(x)} \\\\ \frac{ \frac{d}{dx}\mu(x)}{\mu(x)}=-2\ --\ \textgreater \ \ \frac{d}{dx}(ln(\mu(x)))=-2 \\\\ In(\mu(x))= \int\limits {-2} \, dx \ --\ \textgreater \ \ In(\mu(x))=-2x+c_1 \\\\ \mu(x)=e^{-2x+c_1} \ --\ \textgreater \ \ \| \ \mu(x)=e^{-2x} \ \|$

$Voltando\ para\ a\ EDO: \\\\ \frac{d}{dx}y-2y=e^{2x} \ --\ \textgreater \ \ \mu(x).{[\frac{d}{dx}y-2y]}=\mu(x).e^{2x} \\\\ e^{-2x}.\frac{d}{dx}y-e^{-2x}.2y=e^{-2x}.e^{2x} \\\\ e^{-2x}}. \frac{d}{dx}y-e^{-2x}.2y=1 \ --\ \textgreater \ \ (f.g)'=f'.g+f.g' \\\\ \frac{d}{dx}(e^{-2x}y)=1 \ --\ \textgreater \ \ e^{-2x}y= \int\limits {1} \, dx \\\\ e^{-2x}y=x+c_1 \ --\ \textgreater \ \ \| \ y= \frac{x+c_1}{e^{-2x}} \ \|$

Answer 2

Resposta correta: y = ( x+c)e^2x