Question

Show that, for the differential equation d^2y/dx^2 + a(x)dy/dX + b(x)y=0, e^(mx) is a particular integral if m^2+am+b=0. Hence find the value of m so that e^(mx) is a particular integral of the equation (x-2)d^2y/dx^2-(4x-7)dy/dX + (4x-6)y=0.

Answer 1

ANSWER on Question #86334 – Math – Differential Equations

QUESTION

Show that, for the differential equation

$$\frac{d^2 y}{dx^2} + a(x) \frac{dy}{dx} + b(x) y = 0$$

$e^{mx}$ is a particular integral if $m^2 + am + b = 0$. Hence find the value of $m$ so that $e^{mx}$ is a particular integral of the equation

$$(x - 2) \frac{d^2 y}{dx^2} - (4x - 7) \frac{dy}{dx} + (4x - 6) y = 0$$

SOLUTION

Find the indicated derivatives of the function $y = e^{mx}$ and substitute in the initial equation

$$y = e^{mx} \rightarrow \left\{ \begin{array}{c} \frac{dy}{dx} = \frac{d}{dx}(e^{mx}) = m \cdot e^{mx} \\ \frac{d^2 y}{dx^2} = \frac{d}{dx} \left(\frac{dy}{dx}\right) = \frac{d}{dx}(m \cdot e^{mx}) = m^2 \cdot e^{mx} \end{array} \right.$$

Then,

$$\frac{d^2 y}{dx^2} + a(x) \frac{dy}{dx} + b(x) y = 0 \rightarrow m^2 \cdot e^{mx} + a(x) \cdot m \cdot e^{mx} + b(x) \cdot e^{mx} = 0 \rightarrow$$ $$e^{mx} \cdot (m^2 + a(x) m + b(x)) = 0 \rightarrow \left\{ \begin{array}{c} e^{mx} \neq 0, \quad \forall x \in \mathbb{R}, \forall m \in \mathbb{R} \\ m^2 + a(x) m + b(x) = 0 \end{array} \right.$$

Conclusion,

Hint: 1) There is an error in the condition, since the $a(x)$ and $b(x)$ record implies that $a(x)$ and $b(x)$ are some functions of variable $x$, the form of which is not specified. A note $m^2 + am + b = 0$ suggests that $a$ and $b$ in arbitrary constants.

2) This equation

$$\frac {d ^ {2} y}{d x ^ {2}} + a (x) \frac {d y}{d x} + b (x) y = 0$$

and the substitution $y = e^{mx}$ suggests the Euler-Cauchy equation. But there, it is assumed that $a(x) = a_0x^n$ and $b(x) = b_0x^k$, to come to a similar equation $m^2 + am + b = 0$.

(More information: https://en.wikipedia.org/wiki/Cauchy%E2%80%93Euler_equation)

If this is not done, then in general, the equation $m^2 + a(x)m + b(x) = 0$ is an equation with two variables, which cannot be solved in the literal sense of the word, but can only be used to express the variable $m$ through the variable $x$.

This idea will clearly demonstrate the next part of the task.

$$(x - 2) \frac {d ^ {2} y}{d x ^ {2}} - (4 x - 7) \frac {d y}{d x} + (4 x - 6) y = 0$$

The solution of the equation will be sought as $y = e^{mx}$. Find the indicated derivatives of the function $y = e^{mx}$ and substitute in the initial equation

$$y = e ^ {m x} \rightarrow \left\{\begin{array}{c}\frac {d y}{d x} = \frac {d}{d x} (e ^ {m x}) = m \cdot e ^ {m x}\\\frac {d ^ {2} y}{d x ^ {2}} = \frac {d}{d x} \Bigl (\frac {d y}{d x} \Bigr) = \frac {d}{d x} (m \cdot e ^ {m x}) = m ^ {2} \cdot e ^ {m x}\end{array}\right.$$

Then,

$$\begin{array}{l} (x - 2) \frac {d ^ {2} y}{d x ^ {2}} - (4 x - 7) \frac {d y}{d x} + (4 x - 6) y = 0 \rightarrow \\ (x - 2) m ^ {2} \cdot e ^ {m x} - (4 x - 7) m \cdot e ^ {m x} + (4 x - 6) \cdot e ^ {m x} = 0 \rightarrow \\ e ^ {m x} \cdot \left((x - 2) m ^ {2} - (4 x - 7) m + (4 x - 6)\right) = 0 \rightarrow \\ \left\{ \begin{array}{c} e ^ {m x} \neq 0, \quad \forall x \in \mathbb {R}, \forall m \in \mathbb {R} \\ (x - 2) m ^ {2} - (4 x - 7) m + (4 x - 6) = 0 \end{array} \right. \\ \end{array}$$

Got an equation with two variables $x$ and $m$:

$$(x - 2)m^2 - (4x - 7)m + (4x - 6) = 0$$

This is a quadratic equation for the variable $m$. Let's try to solve it using the discriminant formula

$$\left\{ \begin{array}{c} a = x - 2 \\ b = -(4x - 7) \to D = b^2 - 4ac = \left(-(4x - 7)\right)^2 - 4 \cdot (x - 2) \cdot (4x - 6) \to \\ c = 4x - 6 \end{array} \right.$$ $$D = 16x^2 - 56x + 49 - 4 \cdot (4x^2 - 6x - 8x + 12) \to$$ $$D = 16x^2 - 56x + 49 - 16x^2 + 24x + 32x - 48 = 1 \to \sqrt{D} = \sqrt{1} = 1$$ $$\left\{ \begin{array}{l} m_1 = \frac{-b - \sqrt{D}}{2a} = \frac{4x - 7 - 1}{2(x - 2)} = \frac{4x - 8}{2(x - 2)} = \frac{2(2x - 4)}{2(x - 2)} = \frac{2x - 4}{x - 2} \\ m_1 = \frac{-b + \sqrt{D}}{2a} = \frac{4x - 7 + 1}{2(x - 2)} = \frac{4x - 6}{2(x - 2)} = \frac{2(2x - 3)}{2(x - 2)} = \frac{2x - 3}{x - 2} \end{array} \right.$$

As we see, $m$ is a function of the variable $x$, not a number.

Hint: I can only indicate that you solve the specified equation on the site https://www.wolframalpha.com

That can get a solution

$$y(x) = \frac{C_1 e^{2x - 4} \sqrt{x - 2}}{\sqrt{4 - 2x}} + \frac{C_2 e^{2x - 4} (x - 4) \sqrt{x - 2} x}{\sqrt{2} \sqrt{2 - x}}$$

(https://www.wolframalpha.com/input/?i=(x-2)++(d%5E2+y)%2F(dx%5E2+)-(4x-7)++dy%2Fdx%2B+(4x-6)y%3D0)

We can see that in a machine solution there is a degree $e^{2x - 4}$.

Q.E.D.

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