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Chapter 4: Problem 66

(a) Solve the initial value problem \(6 x^{2} y^{\prime \prime}+\) \(5 x y^{\prime}-y=0, y(1)=a, y^{\prime}(1)=b\). (b) Find conditions on \(a\) and \(b\) so that \(\lim _{x \rightarrow 0^{+}} y(x)=0\). Graph several solutions to confirm your results. (c) Find conditions on \(a\) and \(b\) so that \(\lim _{x \rightarrow \infty} y(x)=0\). Graph several solutions to confirm your results. (d) If both \(a\) and \(b\) are not zero, is it possible to find \(a\) and \(b\) so that both \(\lim _{x \rightarrow 0^{+}} y(x)=0\) and \(\lim _{x \rightarrow \infty} y(x)=0\) ? Explain.

Short Answer

Expert verified
Solve characteristic equation to find general solution. Apply initial conditions, then analyze the limits as x approaches 0 and 鈭� to find conditions. Both conditions can't happen simultaneously unless at least one is zero.

Step by step solution

01

Solve Homogeneous Differential Equation

Consider the given differential equation: \[ 6x^2 y'' + 5xy' - y = 0 \]Assume a solution of the form \( y = x^m \). Substitute it into the equation:\[ 6x^2 m(m-1)x^{m-2} + 5x m x^{m-1} - x^m = 0 \]Simplify to obtain the characteristic equation:\[ 6m(m-1) + 5m - 1 = 0 \]\[ 6m^2 - 6m + 5m - 1 = 0 \]\[ 6m^2 - m - 1 = 0 \]
02

Solve the Characteristic Equation

Solve the characteristic equation for \( m \):\[ 6m^2 - m - 1 = 0 \]Using the quadratic formula \( m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 6 \), \( b = -1 \), and \( c = -1 \):\[ m = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 6 \cdot (-1)}}{2 \cdot 6} \]\[ m = \frac{1 \pm \sqrt{1 + 24}}{12} \]\[ m = \frac{1 \pm 5}{12} \]\[ m_1 = \frac{6}{12} = \frac{1}{2}, \, m_2 = \frac{-4}{12} = -\frac{1}{3} \]
03

Form the General Solution

Combine the solutions into the general form:\[ y(x) = C_1 x^{1/2} + C_2 x^{-1/3} \]Use the initial conditions \( y(1) = a \) and \( y'(1) = b \) to find \( C_1 \) and \( C_2 \).
04

Apply Initial Conditions

Use the initial condition \( y(1) = a \):\[ a = C_1 (1)^{1/2} + C_2 (1)^{-1/3} \]\[ a = C_1 + C_2 \]Differentiate the general solution for \( y' \):\[ y'(x) = \frac{1}{2} C_1 x^{-1/2} - \frac{1}{3} C_2 x^{-4/3} \]Apply \( y'(1) = b \):\[ b = \frac{1}{2} C_1 (1)^{-1/2} - \frac{1}{3} C_2 (1)^{-4/3} \]\[ b = \frac{1}{2} C_1 - \frac{1}{3} C_2 \]Solve these two equations for \( C_1 \) and \( C_2 \).
05

Conditions for Limit as x approaches 0

To find conditions on \( a \) and \( b \) such that \( \lim_{x \rightarrow 0^+} y(x) = 0 \), let鈥檚 analyze each term in the general solution.Since \( x^{-1/3} \) becomes unbounded as \( x \rightarrow 0 \), we need \( C_2 = 0 \).Then the solution simplifies to \( y(x) = C_1 x^{1/2} \), and \( C_1 \) can be any constant. Hence, the condition is:\[ C_2 = 0 \]Using initial conditions, determine the corresponding \( a \) and \( b \). Given \( a = C_1 + C_2 \) and \( b = \frac{1}{2} C_1 - \frac{1}{3} C_2 \), substitute \( C_2 = 0 \):\[ a = C_1 \]\[ b = \frac{1}{2} a \]
06

Conditions for Limit as x approaches 鈭�

To find conditions such that \( \lim_{x \rightarrow \infty} y(x) = 0 \), we need \( C_1 = 0 \).Then the solution simplifies to \( y(x) = C_2 x^{-1/3} \).Using initial conditions:\[ a = C_1 + C_2 \]\[ b = \frac{1}{2} C_1 - \frac{1}{3} C_2 \]Substitute \( C_1 = 0 \):\[ a = C_2 \]\[ b = -\frac{1}{3} a \]
07

Possibility of Both Limits Being Zero

If both \( a \) and \( b \) are not zero, check if we can satisfy both conditions:Condition 1: \( C_2 = 0 \) results in \( a = 0 \) or \( b = 0.5a \).Condition 2: \( C_1 = 0 \) results in \( a = 0 \) or \( b = -\frac{1}{3}a \).It is not possible for both conditions to be true simultaneously unless \( a = 0\), which contradicts the assumption. Hence, finding such \( a \) and \( b \) simultaneously is not possible.
08

Graph Solutions

Graph the solutions \( y(x) = C_1 x^{1/2} + C_2 x^{-1/3} \) for various values of \( C_1 \) and \( C_2 \) to visualize the behavior and confirm the findings.

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Key Concepts

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

Homogeneous Differential Equation
In this exercise, we are dealing with a homogeneous differential equation. A differential equation is termed 'homogeneous' if every term is a function of dependent and independent variables where the sum of the degrees of each term is the same. In our given problem:\[ 6x^2 y'' + 5xy' - y = 0 \]Each term's degree concerning both the variable \( x \) and its derivatives is 2. A homogeneous differential equation often simplifies analysis and allows the application of specific methods, like assuming a solution of the form \( y = x^m \). This initial step usually leads us to formulating a characteristic equation.
Characteristic Equation
To solve our homogeneous differential equation, we convert it to a characteristic equation. By assuming a solution of the form \( y = x^m \), and substituting it back into the differential equation, we get:\[ 6m(m-1)x^{m-2} + 5m x^{m-1} - x^m = 0 \]Since \( x^m eq 0 \), the terms reduce to a quadratic equation in terms of \( m \):\[ 6m^2 - m - 1 = 0 \]Solving this quadratic equation using the quadratic formula \( m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), we find:\[ m_1 = \frac{1}{2} \quad \text{and} \quad m_2 = -\frac{1}{3} \]These roots help in forming the general solution of the differential equation.
Initial Conditions
Initial conditions are vital for determining the specific solution of a differential equation. In our problem, we use initial conditions \( y(1) = a \) and \( y'(1) = b \) to find particular values for constants \( C_1 \) and \( C_2 \) in the general solution. Our general solution takes the form:\[ y(x) = C_1 x^{1/2} + C_2 x^{-1/3} \]Applying the initial conditions:\[ a = C_1 + C_2 \]For the derivative \( y'(x) \):\[ y'(x) = \frac{1}{2} C_1 x^{-1/2} - \frac{1}{3} C_2 x^{-4/3} \]Giving another equation:\[ b = \frac{1}{2} C_1 - \frac{1}{3} C_2 \]Solving these equations together will provide us the specific values of \( C_1 \) and \( C_2 \).
Limit Behavior
Limit behavior helps us understand the behavior of the solution as \( x \) approaches 0 or infinity. For instance, to ensure that \( \lim_{x \rightarrow 0^+} y(x) = 0 \), observe if any term in the solution goes to infinity. In our general solution, \( x^{-1/3} \) becomes unbounded as \( x \rightarrow 0 \), so we set \( C_2 = 0 \). This simplifies to:\[ y(x) = C_1 x^{1/2} \]Ensuring the limit behavior effectively refines our solution space and parameter values. Likewise, for \( \lim_{x \rightarrow \infty} y(x) = 0 \), it is necessary to set \( C_1 = 0 \) so that the term \( x^{1/2} \) does not become unbounded.
General Solution of Differential Equations
The general solution of a differential equation encompasses all possible solutions. For the problem given, the general solution, derived using the roots from the characteristic equation, is:\[ y(x) = C_1 x^{1/2} + C_2 x^{-1/3} \]Applying the initial conditions will help tailor this general solution to a specific solution. For instance, using \( y(1) = a \) and \( y'(1) = b \), you will find specific \( C_1 \) and \( C_2 \). These constants are determined by solving the simultaneous equations produced from the initial conditions and thus give you a concrete form of the solution that fits within the initial parameters presented in the problem.

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Most popular questions from this chapter

\(y^{(4)}-16 y=1\)

\(4 x^{2} y^{\prime \prime}+y=x^{3}, y(1)=1, y^{\prime}(1)=-1\)

\(x^{2} y^{\prime \prime}+x y^{\prime}+y=x^{2}, x<0\)

Let \(x=-e^{t}\). (a) Show that \(\frac{d y}{d x}=\frac{d y}{d t} \frac{d t}{d x}=\) \(\frac{1}{x} \frac{d y}{d t}\) and \(\frac{d^{2} y}{d x^{2}}=\frac{1}{x^{2}}\left(\frac{d^{2} y}{d t^{2}}-\frac{d y}{d t}\right)\). (b) Show that the differential equation \(a x^{2} y^{\prime \prime}+b x y^{\prime}+\) \(c y=f(x)\) is transformed into \(a d^{2} y / d t^{2}+\) \((b-a) d y / d t+c y=f\left(-e^{t}\right)\).

\(x^{2} y^{\prime \prime}-x y^{\prime}+y=0, y(-1)=0, y^{\prime}(-1)=1\)
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