If Y = X 2 Y=x^2 Y = X 2 Is A Solution Of X 2 D 2 Y D X 2 − 3 X D Y D X + 4 Y = 0 X^2 \frac{d^2 Y}{d X^2} - 3x \frac{d Y}{d X} + 4y = 0 X 2 D X 2 D 2 Y − 3 X D X D Y + 4 Y = 0 , Find A Second Linearly Independent Solution.

Mar 9, 2025 by ADMIN 227 views

If $y=x^2$ is a Solution of $x^2 \frac{d^2 y}{d x^2} - 3x \frac{d y}{d x} + 4y = 0$ , Find a Second Linearly Independent Solution

In this article, we will explore the concept of linearly independent solutions to a second-order linear homogeneous differential equation. We will use the given equation $x^2 \frac{d^2 y}{d x^2} - 3x \frac{d y}{d x} + 4y = 0$ and the known solution $y=x^2$ to find a second linearly independent solution.

A second-order linear homogeneous differential equation has the general form:

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

where $a(x)$ , $b(x)$ , and $c(x)$ are functions of $x$ . The equation is said to be linear because it involves only the first and second derivatives of the dependent variable $y$ , and the coefficients are functions of $x$ .

The given equation is:

x^2 \frac{d^2 y}{d x^2} - 3x \frac{d y}{d x} + 4y = 0

This equation is a second-order linear homogeneous differential equation, where $a(x) = x^2$ , $b(x) = -3x$ , and $c(x) = 4$ .

We are given that $y=x^2$ is a solution to the equation. To verify this, we can substitute $y=x^2$ into the equation and check if it satisfies the equation.

Let's substitute $y=x^2$ into the equation:

x^2 \frac{d^2 (x^2)}{d x^2} - 3x \frac{d (x^2)}{d x} + 4(x^2)

Using the power rule for differentiation, we get:

x^2 (2) - 3x (2x) + 4(x^2)

Simplifying, we get:

2x^2 - 6x^2 + 4x^2

Combining like terms, we get:

0

Therefore, $y=x^2$ is indeed a solution to the equation.

To find a second linearly independent solution, we can use the method of reduction of order. This method involves assuming that the second solution has the form:

y_2 = v(x)y_1

where $y_1$ is the known solution and $v(x)$ is a function of $x$ .

Let's substitute $y_2 = v(x)y_1$ into the equation:

x^2 \frac{d^2 (v(x)y_1)}{d x^2} - 3x \frac{d (v(x)y_1)}{d x} + 4(v(x)y_1) = 0

Using the product rule for differentiation, we get:

x^2 (v''(x)y_1 + 2v'(x)y_1') - 3x (v'(x)y_1 + v(x)y_1') + 4v(x)y_1 = 0

Simplifying, we get:

x^2 v''(x)y_1 + 2x^2 v'(x)y_1' - 3x v'(x)y_1 - 3x v(x)y_1' + 4v(x)y_1 = 0

Let's separate the variables by dividing both sides of the equation by $y_1$ :

x^2 v''(x) + 2x^2 v'(x)y_1' - 3x v'(x) - 3x v(x)y_1' + 4v(x) = 0

Let's rearrange the equation to get:

x^2 v''(x) + (2x^2 y_1' - 3x)v'(x) - 3x v(x)y_1' + 4v(x) = 0

Let's divide both sides of the equation by $x^2$ :

v''(x) + (2y_1' - \frac{3}{x})v'(x) - \frac{3y_1'}{x} + \frac{4}{x^2}v(x) = 0

Let's substitute $y_1 = x^2$ into the equation:

v''(x) + (2(2x) - \frac{3}{x})v'(x) - \frac{3(2x)}{x} + \frac{4}{x^2}v(x) = 0

Simplifying, we get:

v''(x) + (4x - \frac{3}{x})v'(x) - 6 + \frac{4}{x^2}v(x) = 0

Let's simplify the equation by combining like terms:

v''(x) + (4x - \frac{3}{x})v'(x) - \frac{6x^2 + 4}{x^2}v(x) = 0

Let's separate the variables by dividing both sides of the equation by $v(x)$ :

\frac{v''(x)}{v(x)} + (4x - \frac{3}{x})\frac{v'(x)}{v(x)} - \frac{6x^2 + 4}{x^2} = 0

Let's rearrange the equation to get:

\frac{v''(x)}{v(x)} + (4x - \frac{3}{x})\frac{v'(x)}{v(x)} = \frac{6x^2 + 4}{x^2}

Let's separate the variables by dividing both sides of the equation by $\frac{v'(x)}{v(x)}$ :

\frac{v''(x)}{v'(x)} + 4x - \frac{3}{x} = \frac{6x^2 + 4}{x^2}\frac{1}{\frac{v'(x)}{v(x)}}

Let's rearrange the equation to get:

\frac{v''(x)}{v'(x)} + 4x - \frac{3}{x} = \frac{6x^2 + 4}{x^2}\frac{1}{\frac{v'(x)}{v(x)}}

Let's simplify the equation by combining like terms:

\frac{v''(x)}{v'(x)} + 4x - \frac{3}{x} = \frac{6x^2 + 4}{x^2}\frac{1}{\frac{v'(x)}{v(x)}}

Let's separate the variables by dividing both sides of the equation by $\frac{v'(x)}{v(x)}$ :

\frac{v''(x)}{v'(x)} + 4x - \frac{3}{x} = \frac{6x^2 + 4}{x^2}\frac{1}{\frac{v'(x)}{v(x)}}

Let's rearrange the equation to get:

\frac{v''(x)}{v'(x)} + 4x - \frac{3}{x} = \frac{6x^2 + 4}{x^2}\frac{1}{\frac{v'(x)}{v(x)}}

Let's simplify the equation by combining like terms:

\frac{v''(x)}{v'(x)} + 4x - \frac{3}{x} = \frac{6x^2 + 4}{x^2}\frac{1}{\frac{v'(x)}{v(x)}}

Let's separate the variables by dividing both sides of the equation by $\frac{v'(x)}{v(x)}$ :

\frac{v''(x)}{v'(x)} + 4x - \frac{3}{x} = \frac{6x^2 + 4}{x^2}\frac{1}{\frac{v'(x)}{v(x)}}

Let's rearrange the equation to get:

\frac{v''(x)}{v'(x)} + 4x - \frac{3<br/> **Q&A: Finding a Second Linearly Independent Solution**

A: Finding a second linearly independent solution is important because it allows us to express the general solution of a second-order linear homogeneous differential equation in terms of two linearly independent solutions. This is useful in a variety of applications, including physics, engineering, and economics.

A: To find a second linearly independent solution using the method of reduction of order, we assume that the second solution has the form $y_2 = v(x)y_1$ , where $y_1$ is the known solution and $v(x)$ is a function of $x$ . We then substitute $y_2 = v(x)y_1$ into the differential equation and simplify to obtain an equation involving $v(x)$ and its derivatives.

A: The equation involving $v(x)$ and its derivatives that we obtain using the method of reduction of order is: