Question

Solve the initial value problem :

d²x/dt² -6dx/dt + 9x = 0 , x(0) =6 ,x(0) = -1 .

Answer 1

Answer on Question #70844 – Math – Differential Equations

Question

Solve the initial value problem:

$$\mathrm{d}^2 x / \mathrm{d}t^2 - 6 \mathrm{d}x / \mathrm{d}t + 9x = 0, \quad x'(0) = 6, \quad x(0) = -1.$$

Solution

$$\mathrm{d}^2 x / \mathrm{d}t^2 - 6 \mathrm{d}x / \mathrm{d}t + 9x = 0$$ $$\lambda^2 - 6\lambda + 9 = 0$$ $$\lambda_1 = \lambda_2 = 3$$ $$x(t) = C_1 e^{3t} + C_2 e^{3t} t$$ $$x(0) = C_1 = -1$$ $$x'(t) = 3C_1 e^{3t} + 3C_2 e^{3t} t + C_2 e^{3t}$$ $$x'(0) = 3C_1 + C_2 = 6 \quad \Rightarrow \quad C_2 = 6 - 3C_1 = 6 + 3 = 9$$ $$x(t) = -e^{3t} + 9e^{3t} t = e^{3t}(9t - 1)$$

Answer: $x(t) = e^{3t}(9t - 1)$.

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