Question

Consider the differential equation
x^3y ''' + 10x^2y '' + 16xy ' − 16y = 0; x, x^−4, x^−4 ln(x), (0, ∞).
Verify that the given functions form a fundamental set of solutions of the differential equation on the indicated interval.
The functions satisfy the differential equation and are linearly independent since
W(x, x^−4, x^−4 ln(x)) =
≠ 0
for
0 < x < ∞.

Form the general solution.
y =


Could someone help me i´ve been on this one for hours I don´t know how to do it and my assignment is due in about an 1 and a half

Answer 1

ANSWER on Question #85725 – Math – Differential Equations

QUESTION

Consider the differential equation

$$x ^ {3} y ^ {\prime \prime \prime} + 10 x ^ {2} y ^ {\prime \prime} + 16 x y ^ {\prime} - 16 y = 0 \rightarrow x, x ^ {- 4}, x ^ {- 4} \ln (x), \quad \forall x \in (0, \infty)$$

Verify that the given functions form a fundamental set of solutions of the differential equation on the indicated interval.

The functions satisfy the differential equation and are linearly independent since

$$W (x, x ^ {- 4}, x ^ {- 4} \ln (x)) \neq 0, \quad \text{for } 0 < x < \infty$$

Form the general solution.

$$y _ {\text{general}} (x) = \dots$$

SOLUTION

**Definition**: for $n$ real- or complex-valued functions $\{f_1, f_2, \ldots, f_n\}$ which are $(n - 1)$ times differentiable on an interval $I$, the Wronskian $W(f_1, f_2, \ldots, f_n)$ as a function on $I$ is defined by

$$W (f _ {1}, f _ {2}, \ldots , f _ {n}) = \left| \begin{array}{c c c c} f _ {1} (x) & f _ {2} (x) & \ldots & f _ {n} (x) \\ f _ {1} ^ {\prime} (x) & f _ {2} ^ {\prime} (x) & \ldots & f _ {n} ^ {\prime} (x) \\ \vdots & \vdots & \ddots & \vdots \\ f _ {1} ^ {(n - 1)} (x) & f _ {2} ^ {(n - 1)} (x) & \ldots & f _ {n} ^ {(n - 1)} (x) \end{array} \right|$$

(More information: https://en.wikipedia.org/wiki/Wronskian)

In our case,

$$W (x, x ^ {- 4}, x ^ {- 4} \ln (x)) \rightarrow \left\{ \begin{array}{c} f _ {1} (x) = x \\ f _ {2} (x) = x ^ {- 4} \\ f _ {3} (x) = x ^ {- 4} \ln (x) \end{array} \right. \rightarrow \left\{ \begin{array}{c} f _ {1} ^ {\prime} (x) = 1 \\ f _ {2} ^ {\prime} (x) = - 4 \cdot x ^ {- 5} \\ f _ {3} ^ {\prime} (x) = - 4 \cdot x ^ {- 5} \ln (x) + x ^ {- 4} \cdot x ^ {- 1} \end{array} \right. \rightarrow$$ $$\left\{ \begin{array}{c} f _ {1} ^ {\prime} (x) = 1 \\ f _ {2} ^ {\prime} (x) = - 4 \cdot x ^ {- 5} \\ f _ {3} ^ {\prime} (x) = x ^ {- 5} \cdot (1 - 4 \ln (x)) \end{array} \right. \to \left\{ \begin{array}{c} f _ {1} ^ {\prime \prime} (x) = 0 \\ f _ {2} ^ {\prime \prime} (x) = (- 4) (- 5) \cdot x ^ {- 6} \\ f _ {3} ^ {\prime \prime} (x) = - 5 \cdot x ^ {- 6} \cdot (1 - 4 \ln (x)) + x ^ {- 5} \cdot (- 4 x ^ {- 1}) \end{array} \right. \to$$ $$\left\{ \begin{array}{c} f _ {1} ^ {\prime \prime} (x) = 0 \\ f _ {2} ^ {\prime \prime} (x) = 20 \cdot x ^ {- 6} \\ f _ {3} ^ {\prime \prime} (x) = x ^ {- 6} (- 5 + 20 \ln (x) - 4) \end{array} \right. \to \left\{ \begin{array}{c} f _ {1} ^ {\prime \prime} (x) = 0 \\ f _ {2} ^ {\prime \prime} (x) = 20 \cdot x ^ {- 6} \\ f _ {3} ^ {\prime \prime} (x) = x ^ {- 6} (- 9 + 20 \ln (x)) \end{array} \right.$$

Conclusion,

$$\left\{ \begin{array}{l} f _ {1} (x) = x \\ f _ {2} (x) = \frac {1}{x ^ {4}} \\ f _ {3} (x) = \frac {\ln (x)}{x ^ {4}} \end{array} \right. \to \left\{ \begin{array}{c} f _ {1} ^ {\prime} (x) = 1 \\ f _ {2} ^ {\prime} (x) = - \frac {4}{x ^ {5}} \\ f _ {3} ^ {\prime} (x) = \frac {1 - 4 \ln (x)}{x ^ {5}} \end{array} \right. \to \left\{ \begin{array}{c} f _ {1} ^ {\prime \prime} (x) = 0 \\ f _ {2} ^ {\prime \prime} (x) = \frac {2 0}{x ^ {6}} \\ f _ {3} ^ {\prime \prime} (x) = \frac {- 9 + 2 0 \ln (x)}{x ^ {6}} \end{array} \right.$$

Then,

W (x, x ^ {- 4}, x ^ {- 4} \ln (x)) = \left| \begin{array}{c c c} x & \frac {1}{x ^ {4}} & \frac {\ln (x)}{x ^ {4}} \\ 1 & - \frac {4}{x ^ {5}} & \frac {1 - 4 \ln (x)}{x ^ {5}} \\ 0 & \frac {2 0}{x ^ {6}} & \frac {- 9 + 2 0 \ln (x)}{x ^ {6}} \end{array} \right| = \\ = x \cdot \left(- \frac {4}{x ^ {5}}\right) \cdot \frac {- 9 + 2 0 \ln (x)}{x ^ {6}} + 1 \cdot \frac {2 0}{x ^ {6}} \cdot \frac {\ln (x)}{x ^ {4}} + 0 \cdot \frac {1}{x ^ {4}} \cdot \frac {1 - 4 \ln (x)}{x ^ {5}} - \\ - 0 \cdot \left(- \frac {4}{x ^ {5}}\right) \cdot \frac {\ln (x)}{x ^ {4}} - x \cdot \frac {2 0}{x ^ {6}} \cdot \frac {1 - 4 \ln (x)}{x ^ {5}} - 1 \cdot \frac {1}{x ^ {4}} \cdot \frac {- 9 + 2 0 \ln (x)}{x ^ {6}} = \\ = \frac {3 6 - 8 0 \ln (x)}{x ^ {1 0}} + \frac {2 0 \ln (x)}{x ^ {1 0}} + - 0 - 0 - \frac {2 0 - 8 0 \ln (x)}{x ^ {1 0}} - \frac {2 0 \ln (x) - 9}{x ^ {1 0}} = \\ = \frac {3 6 - 8 0 \ln (x) + 2 0 \ln (x) - (2 0 - 8 0 \ln (x)) - (2 0 \ln (x) - 9)}{x ^ {1 0}} = \\ = \frac {3 6 - 8 0 \ln (x) + 2 0 \ln (x) - 2 0 + 8 0 \ln (x) - 2 0 \ln (x) + 9}{x ^ {1 0}} = \\ = \frac {(3 6 - 2 0 + 9) + \ln (x) \cdot (- 8 0 + 2 0 + 8 0 - 2 0)}{x ^ {1 0}} = \frac {2 5}{x ^ {1 0}} \\ \end{array} \right.

Conclusion,

$$W (x, x ^ {- 4}, x ^ {- 4} \ln (x)) = \frac {2 5}{x ^ {1 0}} \neq 0 \text{ for } 0 < x < \infty \rightarrow \{x, x ^ {- 4}, x ^ {- 4} \ln (x) \} - \text{linearly independence}$$

Then,

$$y _ {g e n e r a l} (x) = C _ {1} \cdot x + C _ {2} \cdot x ^ {- 4} + C _ {3} \cdot x ^ {- 4} \ln (x) + C _ {4}$$

For verification, we can use

https://www.wolframalpha.com/input/?i=x%5E3*y%27%27%27%2B10*x%5E2*y%27%27%2B16*x*y%27-16y%3D0

ANSWER

$$W(x, x^{-4}, x^{-4} \ln(x)) = \frac{25}{x^{10}} \neq 0 \quad \text{for} \quad 0 < x < \infty \rightarrow \{x, x^{-4}, x^{-4} \ln(x)\} - \text{linearly independence}$$ $$y_{\text{general}}(x) = C_1 \cdot x + C_2 \cdot x^{-4} + C_3 \cdot x^{-4} \ln(x) + C_4$$

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