Question

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Answer 1

Answer on Question #60153 – Math – Differential Equations

Question

Find the general solution of $x^4 y'' + x^3 y' - 4x^2 y = 1, x \in (0, \infty)$, given that $y_1 = x^2$ is a solution of the associated homogeneous equation.

Solution

The associated homogeneous equation is

$$x ^ {4} y ^ {\prime \prime} + x ^ {3} y ^ {\prime} - 4 x ^ {2} y = 0$$

Its solutions can be found in the following form:

$$y = x ^ {n}$$

So

$$x ^ {4} y ^ {\prime \prime} + x ^ {3} y ^ {\prime} - 4 x ^ {2} y = x ^ {4} n (n - 1) x ^ {n - 2} + x ^ {3} n x ^ {n - 1} - 4 x ^ {2} x ^ {n} = (n (n - 1) + n - 4) x ^ {n + 1} = 0$$ $$n (n - 1) + n - 4 = 0 \rightarrow n = 2 \text{ and } n = - 2$$

Thus,

$y_{1} = x^{2}$ and $y_{2} = x^{-2}$ are solutions of (1).

Let $f_{2}(x) = x^{4}, f_{1}(x) = x^{3}, f_{0}(x) = -4x^{2}, g(x) = 1, f_{2}(x)y'' + f_{1}(x)y' + f_{0}(x)y = g(x),$

$$W (x) = y _ {1} (x) \frac {d}{d x} y _ {2} (x) - y _ {2} (x) \frac {d}{d x} y _ {1} (x) = x ^ {2} \frac {d}{d x} (x ^ {- 2}) - x ^ {- 2} \frac {d}{d x} x ^ {2} = - \frac {2 x ^ {2}}{x ^ {3}} - \frac {2 x}{x ^ {2}} = - \frac {4}{x}$$

The general solution of $x^4 y'' + x^3 y' - 4x^2 y = 1$ is

$$\begin{array}{l} y (x) = C _ {1} y _ {1} (x) + C _ {2} y _ {2} (x) + y _ {2} (x) \int y _ {1} (x) \frac {g (x)}{f _ {2} (x)} \frac {d x}{W (x)} - y _ {1} (x) \int y _ {2} (x) \frac {g (x)}{f _ {2} (x)} \frac {d x}{W (x)} = \\ = C _ {1} x ^ {2} + C _ {2} x ^ {- 2} + x ^ {- 2} \int x ^ {2} \frac {1}{x ^ {4}} \frac {d x}{\left(- \frac {4}{x}\right)} - x ^ {2} \int x ^ {- 2} \frac {1}{x ^ {4}} \frac {d x}{\left(- \frac {4}{x}\right)} = \\ = C _ {1} x ^ {2} + \frac {C _ {2}}{x ^ {2}} - \frac {1}{4 x ^ {2}} \int \frac {d x}{x} + \frac {x ^ {2}}{4} \int \frac {d x}{x ^ {5}} = \\ \end{array}$$

$= C_{1}x^{2} + \frac{C_{2}}{x^{2}} -\frac{1}{4x^{2}} \ln |x| + \frac{x^{2}}{4}\cdot \frac{x^{-5 + 1}}{-5 + 1} == C_{1}x^{2} + \frac{C_{2}}{x^{2}} -\frac{1}{4x^{2}} \ln |x| - \frac{1}{16x^{2}} = Cx^{2} + \frac{1}{x^{2}}\left(D - \frac{1}{4}\ln |x|\right)$, where $C,D$ are arbitrary real constants.

Answer: $Cx^{2} + \frac{1}{x^{2}}\left(D - \frac{1}{4}\ln |x|\right)$.

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