1. 5y" + y' = 6x com y(0)=0 e y (0)=-10 Me ajudem a resolver??

Bom dia!Precisa de quantas?

Entrar em contato0. 81 9 9701 1759

\[\eqalign{ & 5{y^{''}} + y^{'} = 6x \cr & y(0) = 0 \cr & y'(0) = - 10 }\]

Encontrar as raízes:

\[\eqalign{ & {y_c} = {c_1}{e^{{r_1}x}} + {c_2}{e^{{r_2}}}x \cr & 5{r^2} + r = r(5r + 1) = 0 \cr & {r^1} = 0 \cr & {r^2} = - {1 \over 5} }\]

O resultado final será:

\[{y_c} = {c_1}{e^{0x}} + {c_2}e - {x \over 5} = {c_1} + {c_2}e - {x \over 5}\]

\[{y_c}(0) = {c_1} + {c_2} = 0\]

\[{c_2} = 50\]

\[{c_1} = - 50\]

\[{y_c} = - 50 + 50{e^{ - {x \over 5}}}\]

\[{y_p} = a{x^2} + bx + c\]

\[{y_p'}=2a{x^2}+ bx + c\]

\[{y_p}^{''} = 2a\]

\[5{y_p}^{''} + {y_p'}= 6x\]

\[10a + 2ax + b = 6x\]

\[\eqalign{ & 10a + b = 0 \cr & 2ax = 6x \cr & a = 3\,\,\,b = - 30 }\]

\[{y_p} = 3{x^2} - 30x\]

Equação resolvida:

\[\]
= - 50 + 50{e^{{ - {x \over 5}}} + 3{x}2