Let y1(x) and y2(x) be two linearly independent solutions of the differential equationx2 y”(x) – 2xy’(x) – 4y(x) = 0 for x ∈[1, 10].Consider the Wronskian W(x) = y1(x)y2’(x) – y2(x)y1’(x). If W(1) = 1, then W(3) – W(2) equalsa)1b)2c)3d)5Correct answer is option 'D'. Can you explain this answer?

Question

Let y1(x) and y2(x) be two linearly independent solutions of the differential equationx2 y&#8221;(x) &#8211; 2xy&#8217;(x) &#8211; 4y(x) = 0 for x &#8712;[1, 10].Consider the Wronskian W(x) = y1(x)y2&#8217;(x) &#8211; y2(x)y1&#8217;(x). If W(1) = 1, then W(3) &#8211; W(2) equalsa)1b)2c)3d)5Correct answer is option 'D'. Can you explain this answer?

Answer

Solution:

Given:
y1(x) and y2(x) are two linearly independent solutions of the differential equation x^2y(x) - 2xy(x) + 4y(x) = 0 for x [1, 10].

We are given the Wronskian W(x) = y1(x)y2(x) - y2(x)y1(x).

To find: W(3) - W(2)

Step 1: Finding the Wronskian W(x)

We know that the Wronskian of two functions is given by the determinant of the matrix containing the functions and their derivatives.

W(x) = | y1(x) y2(x) |
| y1'(x) y2'(x) |

Let's find the derivatives of y1(x) and y2(x) with respect to x.

Differentiating the given differential equation:
x^2y(x) - 2xy(x) + 4y(x) = 0

Taking the derivative of both sides with respect to x:
2xy(x) + x^2y'(x) - 2y(x) - 2xy'(x) + 4y'(x) = 0

Simplifying the equation:
x^2y'(x) - 2xy'(x) + 4y'(x) = 2y(x) - 4y(x)
x^2y'(x) - 2xy'(x) + 4y'(x) = -2y(x)

Therefore, we have the following system of differential equations:
y1'(x) = -2y1(x)/x^2 + 2y1(x)/x - 4y1(x)/x^2
y2'(x) = -2y2(x)/x^2 + 2y2(x)/x - 4y2(x)/x^2

Now, we can find the Wronskian W(x) by substituting the values of y1(x), y2(x), y1'(x), and y2'(x) into the matrix.

W(x) = | y1(x) y2(x) |
| y1'(x) y2'(x) |

W(x) = | y1(x) y2(x) |
| -2y1(x)/x^2 + 2y1(x)/x - 4y1(x)/x^2 -2y2(x)/x^2 + 2y2(x)/x - 4y2(x)/x^2 |

Simplifying further:
W(x) = | y1(x) y2(x) |
| -2y1(x)/x -2y2(x)/x |

Step 2: Finding W(1) and W(3)

Substituting x = 1 into the Wronskian W(x), we get:
W(1) = | y1(1) y2(1) |
| -2y1(1) -2y2(1) |

Given that W(1) = 1, we have:
1 = y1(1) * y2(1) - (-2y1(1)) * (-

Answer

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