Question

Primes denote derivatives with respect to x. y′′ −3y′ +2y=0

Answer 1

The solution to the differential equation
`\(y'' - 3y' + 2y = 0\)`with primes denoting derivatives with respect to
`\(x\) is \(y(x) = c_1e^(2x) + c_2e^x\), where \(c_1\) and \(c_2\)` are arbitrary constants.

Step-by-step explanation

The given differential equation is a linear homogeneous second-order ordinary differential equation with constant coefficients. The characteristic equation is obtained by substituting
`\(y = e^(rx)\)` into the differential equation:

`\[ r^2 - 3r + 2 = 0 \]`

Factoring the quadratic equation yields
`\( (r - 2)(r - 1) = 0 \), so the roots are \( r_1 = 2 \) and \( r_2 = 1 \).`The general solution is a linear combination of the homogeneous solutions:

`\[ y(x) = c_1e^(r_1x) + c_2e^(r_2x) \]`

Substituting the values of
`\( r_1 \) and \( r_2 \), we get \( y(x) = c_1e^(2x) + c_2e^x \). Here, \( c_1 \) and \( c_2 \)` are arbitrary constants determined by initial or boundary conditions if provided.

In summary, the solution to the given differential equation is
`\( y(x) = c_1e^(2x) + c_2e^x \), where \( c_1 \) and \( c_2 \)` are constants that can be determined based on specific conditions of the problem.