Question

Solve the initial value problem: d^2x/dt^2 - 6 dx/dt +9x =0, x(0) =6, X^. (0) =-1

Answer 1

Answer on Question #70070 Math / Differential Equations

Solve the initial value problem: $\frac{d^2x}{dt^2} - 6\frac{dx}{dt} + 9x = 0$, $x(0) = 6$, $x'(0) = -1$.

**Solution:**

Let us consider the initial value problem

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

We will seek for its solution in the form

$$x(t) = e^{mt}.$$

Thus

$$\frac{dx}{dt} = m e^{mt}, \quad \frac{d^2x}{dt^2} = m^2 e^{mt}.$$

Next, we will substitute these expressions into differential equation. We obtain algebraic equation

$$m^2 e^{mt} - 6 m e^{mt} + 9 e^{mt} = 0,$$ $$m^2 - 6m + 9 = 0,$$ $$(m - 3)^2 = 0,$$ $$m_1 = m_2 = 3.$$

So, the solution of the differential equation

$$x(t) = A e^{3t} + B t e^{3t}.$$

The initial conditions give

$$x(0) = A = 6,$$ $$x'(0) = 3A + B = -1, \quad \rightarrow B = -19.$$

Finally

$$x(t) = 6 e^{3t} - 19 t e^{3t}.$$

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

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