Question

Find a general solution to the given differential equation.

2 zʹ'+zʹ-2 z=0
A general solution is z(t)=__

Answer 1

A general solution to the given differential equation
`\(2z'' + z' - 2z = 0\) is \(z(t) = C_1 e^(2t) + C_2 e^(-t)\)`, where
`\(C_1\) and \(C_2\)` are arbitrary constants.

Step-by-step explanation:

Certainly! To find the general solution to the given differential equation
`\(2z'' + z' - 2z = 0\),` we can follow these steps:

1. Assume a Solution:

Assume a solution in the form
`\(z(t) = e^(rt)\), where \(r\)` is a constant.

2. Find Derivatives:

Calculate the first and second derivatives of
`\(z(t)\):`

`\[z'(t) = re^(rt), \quad z''(t) = r^2e^(rt)\]`

3. Substitute into the Differential Equation:

Substitute
`\(z(t)\), \(z'(t)\), and \(z''(t)\)` into the differential equation:

`\[2r^2e^(rt) + re^(rt) - 2e^(rt) = 0\]`

4. Factor Out
`\(e^(rt)\):`

Factor out the common term
`\(e^(rt)\):`

`\[e^(rt)(2r^2 + r - 2) = 0\]`

5. Set the Expression Inside Parentheses to Zero:

`Set \(2r^2 + r - 2 = 0\)` and solve for
`\(r\).` The roots are
`\(r_1 = 2\) and \(r_2 = -1\).`

6. Write Down the General Solution:

The general solution is a linear combination of the fundamental solutions corresponding to the roots:

`\[z(t) = C_1 e^(2t) + C_2 e^(-t)\]`

This is the general solution to the given differential equation, where
`\(C_1\) and \(C_2\)` are arbitrary constants. The solution covers all possible scenarios and initial conditions for the differential equation, providing a versatile representation of the family of solutions.