91Ó°ÊÓ

To solve our given second-order linear homogeneous differential equation, we start with the characteristic equation. By writing the associated characteristic equation for \[10 x'' + \frac{1}{10} x = 0\]we get:\[10 r^2 + \frac{1}{10} = 0\]This characteristic equation helps us find the roots, which are crucial for determining the general solution. Isolating the term containing \(r^2\), we multiply both sides by 10:\[100 r^2 + 1 = 0\]Then, rearranging, we solve for \(r^2\):\[r^2 = -\frac{1}{100}\]Finally, we solve for \(r\):\[r = \boldsymbol{ \frac{\boldsymbol{\text{i}}}{10}}\]These roots give us essential information on how the solutions to our differential equation behave.

When the characteristic equation gives us complex roots, like \(r = \boldsymbol{\text{i}}/10\) in our problem, it indicates an oscillatory nature of the solution. Complex roots typically appear in pairs of the form \( \boldsymbol{\text{α ± iβ}}\). Here, \( \boldsymbol{\text{α}} = 0 \) and \( \boldsymbol{\text{β}} = 1/10\).The general solution for a differential equation with such roots contains both sine and cosine functions. Specifically, we write it as:\[x(t) = e^{\boldsymbol{\text{α}} t} ( C_1 \boldsymbol{\text{cos}}(\boldsymbol{\text{β}} t) + C_2 \boldsymbol{\text{sin}}(\boldsymbol{\text{β}} t) )\]Given \( \boldsymbol{\text{α}} = 0 \) and \( \boldsymbol{\text{β}} = \frac{1}{10} \), this simplifies to:\[x(t) = C_1 \boldsymbol{\text{cos}} ( \frac{t}{10} ) + C_2 \boldsymbol{\text{sin}} ( \frac{t}{10} ) \]This setup allows us to move forward by applying initial conditions.

Initial conditions help us find the specific constants in our general solution. For our problem, we're given:

• \(x(0) = -5\)
• \(x'(0) = 1\)

Starting with \(x(0) = -5\):\[x(0) = C_1 \boldsymbol{\text{cos}}(0) + C_2 \boldsymbol{\text{sin}}(0)\]since \( \boldsymbol{\text{cos}}(0) = 1 \) and \( \boldsymbol{\text{sin}}(0) = 0 \):\[\boldsymbol{-5} = C_1\]Thus, \( C_1 = -5\).To find \( C_2 \), we use \( x'(0) = 1\). First, compute the derivative \( x'(t) \):\[x'(t) = - \frac{1}{10} \boldsymbol{\text{sin}} \bigg( \frac{t}{10} \bigg) C_1 + \frac{1}{10} \boldsymbol{\text{cos}} \bigg( \frac{t}{10} \bigg) C_2 \]At \(t = 0\):\[1 = \frac{1}{10} C_2\]Thus, \(C_2 = 10\). Now we substitute back into the general solution.

Using both constants \( C_1 \) and \( C_2 \) obtained from the initial conditions, we can write the specific solution. Given:

• \( C_1 = -5 \)
• \( C_2 = 10 \)

We substitute these values into our general form:\[ x(t) = -5 \boldsymbol{\text{cos}} ( \frac{t}{10} ) + 10 \boldsymbol{\text{sin}} ( \frac{t}{10} ) \]This is the particular solution to our original differential equation:\[10 x '' + \frac{1}{10} x = 0\]It satisfies both our general solution and the given initial conditions, providing us with a complete and specific answer.