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Chapter 18: Problem 15

Obtain two linearly independent solutions valid for \(x > 0\) unless otherwise instructed. $$ 4 x^{2} y^{\prime \prime}+8 x(x+1) y^{\prime}+y=0. $$

Short Answer

Expert verified
The linearly independent solutions are \(y_1 = x^{-1.5 + \sqrt{2}}\) and \(y_2 = x^{-1.5 - \sqrt{2}}\).

Step by step solution

01

Identify the type of equation

The given differential equation is \(4x^2 y'' + 8x(x+1)y' + y = 0\), which is a second-order linear homogeneous differential equation. It appears to be in Cauchy-Euler form.
02

Assume a solution of the form

For Cauchy-Euler equations, we typically assume a solution of the form \(y = x^m\). This assumption leverages the equation's structure since derivatives of \(x^m\) will yield terms proportional to powers of \(x\).
03

Calculate derivatives

Compute the derivatives: \(y' = mx^{m-1}\) and \(y'' = m(m-1)x^{m-2}\). Substitute these into the original equation.
04

Substitute into the differential equation

Substitute \(y = x^m\), \(y' = mx^{m-1}\), and \(y'' = m(m-1)x^{m-2}\) into the equation to get:\[4x^2[m(m-1)x^{m-2}] + 8x(x+1)[mx^{m-1}] + x^m = 0\]Simplify this to:\[4m(m-1)x^m + 8m(x+1)x^m + x^m = 0\]
05

Factor and simplify

Combine like terms and factor out \(x^m\):\[x^m[4m(m-1) + 8m(x+1) + 1] = 0\]This results in:\[x^m[4m^2 + 4m + 1] = 0\]
06

Solve the auxiliary equation

Set the expression inside the brackets equal to zero, leading to the auxiliary equation:\[4m^2 + 12m + 1 = 0\]Solve this quadratic equation using the quadratic formula \(m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 4\), \(b = 12\), and \(c = 1\).
07

Find roots of the auxiliary equation

The quadratic formula gives:\[m = \frac{-12 \pm \sqrt{144 - 16}}{8}\]Which simplifies to:\[m = \frac{-12 \pm \sqrt{128}}{8}\]\[m = \frac{-12 \pm 8\sqrt{2}}{8}\]\[m = -1.5 \pm \sqrt{2}\]
08

Obtain the general solution

The roots \(m_1 = -1.5 + \sqrt{2}\) and \(m_2 = -1.5 - \sqrt{2}\) provide two linearly independent solutions:\[y_1 = x^{m_1} = x^{-1.5 + \sqrt{2}}\]\[y_2 = x^{m_2} = x^{-1.5 - \sqrt{2}}\]The general solution, which is a linear combination of these independent solutions, is:\[y = C_1 x^{-1.5 + \sqrt{2}} + C_2 x^{-1.5 - \sqrt{2}}\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Linearly Independent Solutions
In differential equations, linearly independent solutions play a crucial role in forming a complete solution for homogeneous linear equations. When dealing with a second-order linear homogeneous equation, we need two solutions that are linearly independent to satisfy the general solution.
  • Two solutions, say \(y_1\) and \(y_2\), are linearly independent if no constant multiple or linear combination of one solution equals the other.
  • For example, \(y_1 = x^{m_1}\) and \(y_2 = x^{m_2}\) from the given problem are linearly independent.
The significance of these solutions being linearly independent is that their linear combination, given as \(y = C_1 y_1 + C_2 y_2\), represents the general solution for the differential equation. Here, \(C_1\) and \(C_2\) are arbitrary constants determined by initial conditions.
Second-Order Linear Homogeneous Equations
Second-order linear homogeneous equations are among the most studied types of differential equations. They are characterized by their second derivative and their linear form.
  • The word "homogeneous" implies that all terms depend linearly on the function \(y\) and its derivatives.
  • The standard form of a second-order linear homogeneous differential equation is \(a(x)y'' + b(x)y' + c(x)y = 0\).
The equation in our exercise, \(4x^2 y'' + 8x(x+1)y' + y = 0\), is a perfect example of such an equation. No constant terms are present, and the coefficients of the derivatives are functions of \(x\), classifying it as a Cauchy-Euler equation due to these characteristics.
Auxiliary Equation
An auxiliary equation is essential in solving second-order linear homogeneous differential equations. This algebraic equation helps to find the solutions to these types of differential equations.
  • The auxiliary equation is derived by assuming a solution of the differential equation in the form \(y = x^m\).
  • Upon substitution, the derivatives of \(y\) turn the original differential equation into the auxiliary equation.
In the given exercise, substituting \(y = x^m\) transformed the problem into the auxiliary equation \(4m^2 + 12m + 1 = 0\). Solving this quadratic equation gives the values of \(m\) that are then used to construct the general solution of the original differential equation.
Quadratic Formula in Differential Equations
The quadratic formula is a powerful tool used to solve the auxiliary equation arising from second-order linear homogeneous differential equations.
  • It is applicable when the auxiliary equation is quadratic, as is commonly the case with Cauchy-Euler equations.
  • The formula is given by \(m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).
For our exercise, the auxiliary equation \(4m^2 + 12m + 1 = 0\) required us to use this formula, where \(a = 4\), \(b = 12\), and \(c = 1\), to find the roots: \(m = -1.5 \pm \sqrt{2}\).
This provides the exponents for the linearly independent solutions \(y_1\) and \(y_2\), which yields a complete solution to the original differential equation.

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Most popular questions from this chapter

Find two linearly independent solutions, valid for \(x > 0,\) unless otherwise instructed. $$ x y^{\prime \prime}+(3+2 x) y^{\prime}+8 y=0. $$

In each exercise, obtain solutions valid for \(x > 0\). $$ x^{2} y^{\prime \prime}+x y^{\prime}-\left(x^{2}+4\right) y=0. $$

In each exercise, obtain solutions valid for \(x > 0\). $$ 2 x^{2} y^{\prime \prime}-x(1+2 x) y^{\prime}+(1+3 x) y=0. $$

In each exercise, obtain solutions valid for \(x > 0\). $$ x\left(1-x^{2}\right) y^{\prime \prime}+5\left(1-x^{2}\right) y^{\prime}-4 x y=0. $$

Obtain the general solution near \(x=0\) except when instructed otherwise. State the region of validity of each solution. $$ x(1-x) y^{\prime \prime}-3 y^{\prime}+2 y=0. $$
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