Question

For the homogeneous D.E. x2y′′−2xy′+(x2+2)y=0;y1​=xsin(x) and y2​=xcos(x) are both solutions.

Answer 1

For the homogeneous differential equation
`\(x^2y'' - 2xy' + (x^2 + 2)y = 0\)`, the solutions
`\(y_1 = x\sin(x)\)` and
`\(y_2 = x\cos(x)\)` are linearly independent solutions forming a fundamental set.

Step-by-step explanation:

The given differential equation is a second-order linear homogeneous differential equation with variable coefficients. The solutions
`\(y_1\)` and
`\(y_2\)` are both functions of x and satisfy the differential equation. To determine whether these solutions are linearly independent, we can use the Wronskian determinant.

The Wronskian determinant for two functions
`\(y_1\)` and
`\(y_2\)` is given by:

`\[ W[y_1, y_2] = \begin{vmatrix} y_1 & y_2 \\ y_1' & y_2' \end{vmatrix} \]`

Calculate the Wronskian determinant for
`\(y_1 = x\sin(x)\)` and
`\(y_2 = x\cos(x)\)`. If
`\(W \\eq 0\)`, then
`\(y_1\)` , and
`\(y_2\)` are linearly independent, forming a fundamental set of solutions for the homogeneous differential equation.

In this case, the Wronskian determinant is non-zero, indicating that
`\(y_1 = x\sin(x)\)` and
`\(y_2 = x\cos(x)\)` are linearly independent solutions. Therefore, they form a fundamental set of solutions for the given homogeneous differential equation. This is a crucial result in solving such equations, as it ensures the general solution can be expressed as a linear combination of these independent solutions.