Question

Solve the following Cauchy Euler equation by the method of variation of parameters.
x^2 y^n - xy'+y=2x

Determine the singular points of the following differential equation and classify each singular point as regular or irregular.

(x^2 - 9)^2 y^n + ( x+3) y' +2y = 0

Answer 1

Answer on Question #44251 – Math - Differential Calculus | Equations

Solve the following Cauchy Euler equation by the method of variation of parameters.

$$x ^ {2} y ^ {n} - x y ^ {\prime} + y = 2 x$$

Determine the singular points of the following differential equation and classify each singular point as regular or irregular.

$$(x ^ {2} - 9) ^ {2} y ^ {n} + (x + 3) y ^ {\prime} + 2 y = 0.$$

**Solution.**

Solve the Cauchy Euler equation by the method of variation of parameters.

By definition a Cauchy-Euler equation of order $n$ has the form

$$a _ {n} x ^ {n} y ^ {(n)} (x) + a _ {n - 1} x ^ {n - 1} y ^ {(n - 1)} (x) \dots + a _ {0} y (x) = 0$$

So the correct equation

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

Solve this equation. The general solution will be the sum of the complementary solution and particular solution.

Find the complementary solution by solving $x^{2}y^{\prime \prime} - xy^{\prime} + y = 0$.

Assume a solution to this Euler-Cauchy equation will be proportional to $x^{\lambda}$ for some constant $\lambda$.

Substitute $y(x) = x^{\lambda}$ into the differential equation:

$$x ^ {2} \left(x ^ {\lambda}\right) ^ {\prime \prime} - x \left(x ^ {\lambda}\right) ^ {\prime} + x ^ {\lambda} = 0$$ $$\lambda^ {2} x ^ {\lambda} - 2 \lambda x ^ {\lambda} + x ^ {\lambda} = 0$$

Factor out $x^{\lambda}$:

$$(\lambda^ {2} - 2 \lambda + 1) x ^ {\lambda} = 0$$

Assuming $x \neq 0$, the zeros must come from the polynomial:

$$\lambda^ {2} - 2 \lambda + 1 = 0$$

Solve for $\lambda$:

$$\lambda_ {1, 2} = 1$$

The multiplicity of the root $\lambda = 1$ is 2 which gives $y_{1}(x) = c_{1}x$, $y_{2}(x) = c_{2}x \ln x$ as solutions, where $c_{1}$ and $c_{2}$ are arbitrary constants.

The complementary solution is the sum of the above solutions:

$$y _ {c} (x) = y _ {1} (x) + y _ {2} (x) = c _ {1} x + c _ {2} x \ln x$$

Determine the particular solution by variation of parameters.

List the basis solutions in $y_{c}(x)$:

$$y _ {b _ {1}} (x) = x \text{ and } y _ {b _ {2}} (x) = x \ln x$$

Compute the Wronskian of $y_{b_1}(x)$ and $y_{b_2}(x)$:

$$W (x) = \left| \begin{array}{cc} x & x \ln x \\ \frac {d}{d x} (x) & \frac {d}{d x} (x \ln x) \end{array} \right| = \left| \begin{array}{cc} x & x \ln x \\ 1 & \ln x + 1 \end{array} \right| = x$$

Divide the differential equation by the leading term's coefficient $x^2$:

$$y ^ {\prime \prime} - \frac {y ^ {\prime}}{x} + \frac {y}{x ^ {2}} = \frac {2}{x}$$

Let $f(x) = \frac{2}{x}$. Let $\nu_{1}(x) = -\int \frac{f(x)y_{b_{2}}(x)}{W(x)} dx$ and $\nu_{2}(x) = -\int \frac{f(x)y_{b_{1}}(x)}{W(x)} dx$.

The particular solution will be given by:

$$y _ {p} (x) = v _ {1} (x) y _ {b _ {1}} (x) + v _ {2} (x) y _ {b _ {2}} (x)$$

Compute $\nu_{1}(x)$:

$$\nu_ {1} (x) = - \int \frac {2 \ln x}{x} d x = - \ln^ {2} x$$

Compute $\nu_{2}(x)$:

$$\nu_ {2} (x) = \int \frac {2}{x} d x = 2 \ln x$$

The particular solution is thus:

$$y _ {p} (x) = v _ {1} (x) y _ {b _ {1}} (x) + v _ {2} (x) y _ {b _ {2}} (x) = - x \ln^ {2} x + 2 x \ln^ {2} x$$

Simplify:

$$y _ {p} (x) = x \ln^ {2} x$$

So the general solution:

$$y (x) = y _ {c} (x) + y _ {p} (x) = c _ {1} x + c _ {2} x \ln x + x \ln^ {2} x$$

Answer: $y(x) = c_{1}x + c_{2}x\ln x + x\ln^{2}x$.

Determine the singular points of the following differential equation and classify each singular point as regular or irregular.

Equation in standard form

$$N(x) y'' + P(x) y' + Q(x) y = 0$$

So our equation

$$(x^2 - 9)^2 y'' + (x + 3) y' + 2 y = 0$$

The singular points of the equation are then the points where $N(x) = 0$:

$$(x^2 - 9)^2 = 0$$ $$x_{1,2} = 3, x_{3,4} = -3$$

Suppose $x_0$ is a singular point. Multiplying through by $(x - x_0)^2 / N(x)$, we may rewrite equation as

$$(x - x_0)^2 y'' + (x - x_0) u(x) y' + v(x) y = 0,$$

where

$$u(x) = \frac{(x - x_0) P(x)}{N(x)}, \quad v(x) = \frac{(x - x_0)^2 Q(x)}{N(x)},$$

We say that $x_0$ is a regular singular point if the rational functions $u(x)$ and $v(x)$ have no singularity at $x_0$ – that is, if the factors of $x - x_0$ in $N(x)$ that cause $N(x)$ to vanish at $x_0$ are canceled by such factors in $(x - x_0)P(x)$ and $(x - x_0)^2 Q(x)$. Otherwise $x_0$ is an irregular singular point.

Consider $x_0 = 3$. Find $u(x)$ and $v(x)$:

$$u(x) = \frac{(x - 3)(x + 3)}{(x^2 - 9)^2} = \frac{1}{x^2 - 9}$$ $$v(x) = \frac{(x - 3)^2 \cdot 2}{(x^2 - 9)^2} = \frac{2}{(x + 3)^2}$$

$u(x)$ has singularity at $x_0$, so $x_0 = 3$ is an irregular singular point.

Consider $x_0 = -3$. Find $u(x)$ and $v(x)$:

$$u(x) = \frac{(x + 3)(x + 3)}{(x^2 - 9)^2} = \frac{1}{(x - 3)^2}$$ $$v(x) = \frac{(x + 3)^2 \cdot 2}{(x^2 - 9)^2} = \frac{2}{(x - 3)^2}$$

$u(x)$ and $v(x)$ have no singularity at $x_0$, so $x_0 = -3$ is a regular singular point.

**Answer**: $x_0 = 3$ is an irregular singular point, $x_0 = -3$ is a regular singular point.

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