Question

$Resolva\ :\\\\\\y''\ +y' -\ 3y\ =\ 0\\\\\\com\ \ \ \ y'(0)\ =\ 3\ \ \ \ e\ \ \ \ y(0)\ =\ 1\ \ \ .$

Answer 1

y''+y'-3y=0 , y'(0)=3 , y(0)=1

equação caracteristica
y'' = r² , y'=r , y=1

$r^2 +r -3 = 0 \\\\ \Delta = -1^3- 4*1*(-3) = 13$

a solução vai ser dada por:
$y(x) = C_1*e^{r_1*x}+C_2*e^{r_2*x}$

onde r1 e r2 são as raízes:
$r^2 +r -3 = 0\to \Bmatrix r_1= \frac{-1- \sqrt{13} }{2} \\ r_2 = \frac{-1+ \sqrt{13} }{2} \end$


$y(x)=C_1*e^{r_1*x}+C_2*e^{r_2*x}\\\\y'(x)=C_1*e^{r_1*x}* r_1 + C_2*e^{r_2*x}* r_2 \\ \boxed{\boxed{y'(0)=C_1*r_1 + C_2*r_2=3}}\\\\\\ y''(x)=C_1*e^{r_1*x}*(r_1)^2+C_2*e^{r_2*x}*(r_2)^2\\\\\boxed{\boxed{y''(0)=C_1*(r_1)^2+C_2*(r_2)^2= 1}}$

isolando c1 na primeira equação e substituindo na segunda:

$\boxed{\boxed{C_1= \frac{3- C_2*r_2}{r_1} }} \\\\\\ \frac{3- C_2*r_2}{r_1}*(r_1)^2+C_2*(r_2)^2= 1 \\\\ (3-C_2*r_2)*r_1+C_2*(r_2)^2=1\\\\ 3r_1 -C_2*r_2*r_1+C_2*(r_2)^2=1\\\\C_2*(-r_2*r_1+(r_2)^2)= 1-3r_1 \\\\\boxed{\boxed{C_2= \frac{1-3r_1}{-r_2*r_1+(r_2)^2} }}$

 resolvendo:
$C_2= \frac{1-3* \frac{-1- \sqrt{13} }{2} }{(- \frac{-1+ \sqrt{13} }{2}* \frac{-1- \sqrt{13} }{2}+( \frac{-1+ \sqrt{13} }{2})^2} \approx 1,6836\\\\\\ C_1= \frac{3- 1,6836*( \frac{-1+ \sqrt{13} }{2})}{ \frac{-1- \sqrt{13} }{2}} \approx -0,3503$


$\boxed{\boxed{y(x)=-0,3503*e^{(\frac{-1- \sqrt{13} }{2})*x}+1,6836*e^{(\frac{-1+ \sqrt{13} }{2})*x}}}$


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