Question

Detemine a solução geral homogênea da equação diferencial y"+2y'-8y =0.
Escolha uma:
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Answer 1

$\Large\boxed{\begin{array}{l}\sf y''+2y'-8y=0\\\sf y=e^{mx}\\\sf y'=me^{mx}\\\sf y''=m^2e^{mx}\\\sf m^2e^{mx}+2me^{mx}-8e^{mx}=0\\\sf e^{mx}(m^2+2m-8)=0\\\sf e^{mx}\ne 0\\\sf m^2+2m-8=0\\\sf\Delta=b^2-4ac\\\sf\Delta=2^2-4\cdot1\cdot(-8)\\\sf\Delta=4+32\\\sf\Delta=36\\\sf m=\dfrac{-b\pm\sqrt{\Delta}}{2a}\\\\\sf m=\dfrac{-2\pm\sqrt{36}}{2\cdot1}\\\\\sf m=\dfrac{-2\pm6}{2}\begin{cases}\sf m_1=\dfrac{-2+6}{2}=\dfrac{4}{2}=2\\\\\sf m_2=\dfrac{-2-6}{2}=-\dfrac{8}{2}=-4\end{cases}\end{array}}$

$\Large\boxed{\begin{array}{l}\sf\red{ y_h=C_1e^{2x}+C_2e^{-4x}}\blue{\checkmark}\end{array}}$

Answer 2

Resposta:

Yh(x)=C1e^2x+C2e^-4xC

Explicação passo a passo:

Corrido pelo AVA