Question

Solve the following differential equation for its general solution. It is given that two linearly independent solutions are e−3x and xe−3x. (Do not use decimal fractions anywhere.) fy(4)+(6f+m)y(3)+(9f+l+6m)y′′+(9m+6l)y′+9ly=0 Where, f,m, and l are the number of letters in your first name, middle name, and last name respectively.

Answer 1

The general solution to the given differential equation is
`\(y(x) = c_1 e^(-3x) + c_2 x e^(-3x)\)`, where
`\(c_1\) and \(c_2\)` are arbitrary constants.

Step-by-step explanation:

The provided differential equation is a linear homogeneous second-order differential equation with constant coefficients. The general form of such an equation is given by
`\(ay'' + by' + cy = 0\)`, where
`\(ay'' + by' + cy = 0\)`are constants. In this case, the coefficients are determined by the counts of letters in my first name
`(\(f\))`, middle name
`(\(m\))`, and last name
`(\(l\))`.

The characteristic equation associated with the given differential equation is obtained by substituting
`\(y(x) = e^(rx)\)` into the equation, resulting in
`\(r^2 + (6 + 9f)r + (9l + 6m + 9) = 0\)`. Solving this quadratic equation yields two distinct roots
`\(r_1\)` and
`\(r_2\)`.

Since the given solutions,
`\(e^(-3x)\)` and
`\(xe^(-3x)\`), are linearly independent, the general solution is expressed as a linear combination of these solutions, i.e.,
`\(y(x) = c_1 e^(r_1x) + c_2 e^(r_2x)\`). Substituting the roots into this form and simplifying, we obtain the final general solution
`\(y(x) = c_1 e^(-3x) + c_2 x e^(-3x)\`). The arbitrary constants
`\(c_1\) and \(c_2\)` are determined by any initial conditions or boundary values given for the specific problem at hand.