Question

$Dada\ a\ equac\~ao\ :\\\\\\y''\ +y'\ -\ 6y\ =\ 0\\\\\\ Suas\ soluc\~oes \ atendem\ :\\\\para\ x\ =\ 0\ , \ y\ =\ 1\\para\ x\ =0\ ,\ y'\ =\ 0\\\\\\\\Ent\~ao,\ qual\ sera\ o\ valor\ dessa\ soluc\~ao\ para\ x\ =\ 1\ ?$

Answer 1

$y''+y'-6=0\\y''+y'=6\\\\\text{resolvendo a parte homogenea}\\\\ y''+y'=0\\r^2+r=0\\ r(r+1)=0 \to \bmatrix r_1=0\\r_2=-1\end$

$y_h= C_1*e^{r_1*x}+C_2*e^{r_2*x}\\\\\boxed{\boxed{y_h=C_1+C_2*e^{-x}}}\\ \\\\\text{solucao particular}\\\\ y_p=Ax^2+Bx+C\\(y_p)''=2Ax+B\\(y_p)''=2A\\\\ (y_p)''+(y_p)'= 6\\ 2Ax+B+2A = 6 \to \bmatrix A=0 \\ B=6 \end$


$\boxed{\boxed{y_p= 6x}}\\\\ y(x)=y_h+y_p\\\\ y(x)=C_1+C_2*e^{x}+6x\\\\\\ y(0)= C_1+C_2=1 \to \boxed{\boxed {C_1=1-C_2}}\\\\ y'(x)=-C_2*e^{-x}+6 \\y'(0)=-C_2+6=0 \to \boxed{\boxed{C_2=6}} \to \boxed{\boxed{C_1=-5}}\\\\\\\\y(x)=-5+6e^{-x}+6x\\\\y(1)=-5+6*e^{-1}+6\\\\y(1)=1+ \frac{6}{e}= \frac{e+6}{6}\approx 3,2$

Answer 2

Olá


EDO 2ª ORDEM... PVI.


y'' + y' - 6y = 0

Equação característica.

E.C.: k² + k - 6 = 0

Δ = 25

k1 = -3
k2 = 2


$\mathsf{y = C_1e^{k_1x}~+~C_2e^{k_2x}}\\\\\\\mathsf{y = C_1e^{-3x}~+~C_2e^{2x}}$



PVI


P/ y

Para x = 0, y = 1


P/ y'

Para x = 0, y' = 0



Encontrando y'


$\mathsf{y = C_1e^{-3x}~+~C_2e^{2x}}\\\\\\\mathsf{y'=-3C_1e^{-3x}~+~2C_2e^{2x}}$




Substituindo em y'

y'(0) = 0

$\mathsf{y'=-3C_1e^{-3x}~+~2C_2e^{2x}}\\\\\\\mathsf{0=-3C_1e^{-3\cdot(0)}~+~2C_2e^{2\cdot(0)}}\\\\\\\boxed{\mathsf{-3C_1+2C_2=0}}\qquad\qquad\qquad\longleftarrow\text{Guarde essa informacao}$



Substituindo em y

y(0) = 1


$\mathsf{1 = C_1e^{-3\cdot (0)}~+~C_2e^{2\cdot (0)}}\\\\\\\boxed{\mathsf{C_1+C_2=1}}$



Montando um sistema 2x2 com as informações obtidas no PVI


$\displaystyle \mathsf{ \left \{ {{C_1+C_2=1} \atop {\underline{-3C_1+2C_2=0}}} \right. }\\\\\quad\boxed{\mathsf{C_1= \frac{2}{5} }}$


Encontrando C2


$\displaystyle \mathsf{C_1+C_2=1}\\\\\\\mathsf{ \frac{2}{5}+C_2=1 }\\\\\\\boxed{\mathsf{C_2= \frac{3}{5} }}$



Portanto, y fica sendo


$\displaystyle \boxed{\mathsf{y= \frac{2}{5}e^{-3x}~+~ \frac{3}{5}e^{2x} }}$


Quando x = 1, y = ?


Substituindo quando x = 1


$\displaystyle \mathsf{y(1)= \frac{2}{5}e^{-3\cdot (1)}~+~ \frac{3}{5}e^{2\cdot(1)} }\\\\\\\mathsf{y(1)= \frac{2e^{-3}+3e^2}{5} }\\\\\\\boxed{\mathsf{y(1)\approx 4,45}}$