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Consider the second order equation y\" - 5y\' + 6y = 0. Determine the squareroot

ID: 3006266 • Letter: C

Question

Consider the second order equation y" - 5y' + 6y = 0. Determine the squareroots of the characteristic equation. Find the general solution in terms of real functions. Find a particular solution that satisfies the initial conditions y(0) = -2, y'(0) = 3. Give the corresponding dynamical system, including initial conditions. Classify the equilibrium solution of the dynamical system as to type and stability (no sketch required). Give the general solution to the nonhomogeneous equation y" - 5y' + 6y = 3e^2t.

Explanation / Answer

a) characterisic equation is

s2 - 5s +6 = 0

so roots are 2 & 3.

b) Hence the general solution is

  y (x) = c1 * e2x + c2 * e3x

c) y(0) = -2 y'(0) = 3

c1 + c2 = -2

2c1 +3c2 = 3

c1 = -9 c2 = 7

y (x) = -9 e2x +7 e3x  

d )& e) do it yourself ,if you have doubt ,comment ,

f) the general solution to the non-homogeneous equation is

y(t) = c1 *e2t+ c2* e3t -3 e 2t *t

y * '= the A (2t + 1) e^(2t), Y *' ' = A (4t + 4) e^(2t)

is substituted into the original equation, too: ...... A = -3, y * = -3 te^(2t)

the original equation general solution: y = C1*e^(2t ) + C2*e^(3t) -3*te^(2t)