Question

Find the general solution of the differential equation \( x^{2} y^{\prime \prime}+4 x y^{\prime}+4 y=0 \).

Answer 1

The general solution of the differential equation
`\( x^(2) y^(\prime \prime)+4 x y^(\prime)+4 y=0 \) is given by \( y(x) = c_(1) x^(-2) + c_(2) x^(-2) \ln(x) \), where \( c_(1) \) and \( c_(2) \)`are arbitrary constants.

Step-by-step explanation:

This second-order linear homogeneous differential equation is in the form
`\( x^(2) y^(\prime \prime)+4 x y^(\prime)+4 y=0 \), where \( y^(\prime \prime) \)` denotes the second derivative of
`\( y \) with respect to \( x \).` To find the general solution, we assume a solution of the form
`\( y(x) = x^(r) \), where \( r \)` is a constant. Taking the first and second derivatives, we substitute them into the original equation and solve for
`\( r \).`

The characteristic equation obtained is
`\( r(r+1) = 0 \), yielding \( r = 0 \)`and
`\( r = -1 \).` Therefore, the general solution is a linear combination of the two linearly independent solutions:
`\( y_(1)(x) = c_(1)x^(0) \) and \( y_(2)(x) = c_(2)x^(-1) \)`. However, since
`\( r = 0 \)` is a repeated root, we multiply
`\( y_(1)(x) \) by \( \ln(x) \)` to ensure linear independence. Thus, the general solution is
`\( y(x) = c_(1) x^(-2) + c_(2) x^(-2) \ln(x) \), where \( c_(1) \) and \( c_(2) \)` are arbitrary constants representing the degrees of freedom in the solution space.