Question

d^2x/dt^2 -6dx/dt +9x = 0, x(0) = 6, x(0) = -1

Answer 1

Answer on Question # 71937, Math / Differential Equations

Question

$$\mathrm{d}^2 \mathrm{x}/\mathrm{dt}^2 - 6 \mathrm{dx}/\mathrm{dt} + 9 \mathrm{x} = 0, \quad \mathrm{x}(0) = 6, \quad \mathrm{x}(0) = -1$$

**Solution.** We have the initial value problem

$$\frac{d^2 x}{dt^2} - 6 \frac{dx}{dt} + 9x = 0, \quad x(0) = 6, \quad x(0) = -1$$

First we find the general solution of this differential equation.

Seek a solution in the form $x = e^{mt}$. After substitution of $\frac{dx}{dt} = m e^{mt}$ and $\frac{d^2 x}{dt^2} = m^2 e^{mt}$, equation becomes

$$m^2 e^{mt} - 6m e^{mt} + 9 e^{mt} = 0$$

or

$$e^{mt} (m^2 - 6m + 9) = 0.$$

Since $e^{mt} \neq 0$ the last equation is satisfied only when $m$ is a solution of the algebraic equation

$$m^2 - 6m + 9 = 0.$$

We get $m_1 = m_2 = 3$. Since $m_1 = m_2$ we obtain only one exponential solution

$$x_1(t) = e^{3t}.$$

The second solution is

$$x_2(t) = t e^{3t}.$$

The general solution of the original equation is

$$x(t) = c_1 x_1(t) + c_2 x_2(t) = c_1 e^{3t} + c_2 t e^{3t}$$

where $c_1$ and $c_2$ are constants.

Now we find $c_1$ and $c_2$ using the initial conditions

$$x(0) = c_1 = 6$$ $$x(0) = c_1 = -1$$

However, the constant cannot simultaneously be equal to different numbers, so the initial value problem has no solutions

**Answer:** this initial problem has no solutions.

Addition. If we assume that there is a misprint in the condition, that is, the condition is

$$\frac {d ^ {2} x}{d t ^ {2}} - 6 \frac {d x}{d t} + 9 x = 0, \quad x (0) = 6, \quad x ^ {\prime} (0) = - 1$$

then we can find $c_{1}$ and $c_{2}$

$$x (0) = c _ {1} + 0 = 6 \Rightarrow c _ {1} = 6$$ $$x ^ {\prime} (0) = 3 c _ {1} + c _ {2} = - 1 \Rightarrow c _ {2} = - 1 9$$

Finally

$$x (t) = 6 e ^ {3 t} - 1 9 t e ^ {3 t}$$

Answer: $x(t) = 6e^{3t} - 19te^{3t}$

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