Respuesta :
Answer:
Since 2 and -1 are eigenvalues of the differential equation,
[tex]x(t) = c_{1}e^{-t} + c_{2}e^{2t}[/tex]
is a solution to the differential equation
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The solution to the initial value problem is:
[tex]x(t) = \frac{20}{3}e^{-t} + \frac{10}{3}e^{2t}[/tex]
Step-by-step explanation:
We have the following differential equation:
[tex]x'' - x' - 2x = 0[/tex]
The first step is finding the eigenvalues for this differential equation, that is, finding the roots of the following second order equation:
[tex]r^{2} - r - 2 = 0[/tex]
[tex]\bigtriangleup = (-1)^{2} -4*1*(-2) = 1 + 8 = 9[/tex]
[tex]r_{1} = \frac{-(-1) + \sqrt{\bigtriangleup}}{2*1} = \frac{1 + 3}{2} = 2[/tex]
[tex]r_{2} = \frac{-(-1) - \sqrt{\bigtriangleup}}{2*1} = \frac{1 - 3}{2} = -1[/tex]
Since 2 and -1 are eigenvalues of the differential equation,
[tex]x(t) = c_{1}e^{-t} + c_{2}e^{2t}[/tex]
is a solution to the differential equation.
Solution of the initial value problem:
[tex]x(t) = c_{1}e^{-t} + c_{2}e^{2t}[/tex]
[tex]x(0) = 10[/tex]
[tex]10 = c_{1}e^{-0} + c_{2}e^{2*0}[/tex]
[tex]c_{1} + c_{2} = 10[/tex]
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[tex]x'(t) = -c_{1}e^{-t} + 2c_{2}e^{2t}[/tex]
[tex]x'(0) = 0[/tex]
[tex]0 = -c_{1}e^{-0} + 2c_{2}e^{2*0}[/tex]
[tex]-c_{1} + 2c_{2} = 0[/tex]
[tex]c_{1} = 2c_{2}[/tex]
So, we have to solve the following system:
[tex]c_{1} + c_{2} = 10[/tex]
[tex]c_{1} = 2c_{2}[/tex]
[tex]2c_{2} + c_{2} = 10[/tex]
[tex]3c_{2} = 10[/tex]
[tex]c_{2} = \frac{10}{3}[/tex]
[tex]c_{1} = 2c_{2} = \frac{20}{3}[/tex]
The solution to the initial value problem is:
[tex]x(t) = \frac{20}{3}e^{-t} + \frac{10}{3}e^{2t}[/tex]
