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Chapter 4: Problem 6

\(9 x^{2} y^{\prime \prime}-9 x y^{\prime}+10 y=0\)

Short Answer

Expert verified
The general solution is: \[y = x \left[C_1 \cos\left( \frac{1}{3} \ln x \right) + C_2 \sin\left( \frac{1}{3} \ln x \right) \right]\]

Step by step solution

01

Identify the Type of Differential Equation

This is a second-order linear homogeneous differential equation with variable coefficients: \[9x^2 y'' - 9x y' + 10y = 0\]
02

Simplify the Equation

Divide the entire equation by 9 to obtain: \[x^2 y'' - x y' + \frac{10}{9} y = 0\]
03

Assumption for Solution Form

Assume the solution is of the form: \[y = x^m\] Differentiate to find \(y'\) and \(y''\): \[y' = mx^{m-1}\] \[y'' = m(m-1)x^{m-2}\]
04

Substitute Into the Original Equation

Substituting \(y\), \(y'\), and \(y''\) into the simplified differential equation: \[x^2 \cdot m(m-1)x^{m-2} - x \cdot mx^{m-1} + \frac{10}{9} x^m = 0\] Simplify: \[m(m-1)x^m - m x^m + \frac{10}{9} x^m = 0\]
05

Combine Like Terms

Factor \(x^m\) out: \[x^m \left[m(m-1) - m + \frac{10}{9} \right] = 0\] Therefore, we have: \[m(m-1) - m + \frac{10}{9} = 0\]
06

Solve the Quadratic Equation

Simplify the quadratic expression: \[m^2 - m - m + \frac{10}{9} = 0\] \[m^2 - 2m + \frac{10}{9} = 0\] Multiply by 9 to clear the fraction: \[9m^2 - 18m + 10 = 0\] Solve the quadratic equation using the quadratic formula: \[m = \frac{18 \pm \sqrt{(18)^2 - 4 \cdot 9 \cdot 10}}{2 \cdot 9} = \frac{18 \pm \sqrt{324 - 360}}{18} = \frac{18 \pm \sqrt{-36}}{18} = \frac{18 \pm 6i}{18}\] Simplify: \[m = 1 \pm \frac{1}{3}i\]
07

Write the General Solution

The general solution for the differential equation is: \[y = x^1 \left[C_1 \cos\left( \frac{1}{3} \ln x \right) + C_2 \sin\left( \frac{1}{3} \ln x \right) \right]\]

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Key Concepts

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

Assumption for Solution Form
To solve a second-order linear homogeneous differential equation like \[9x^2 y'' - 9x y' + 10y = 0\] we assume the solution takes a specific form. For this case, we use the assumption \[y = x^m\]. This is because the equation has variable coefficients involving powers of x. By differentiating, we find:\[y' = mx^{m-1}\] and\[y'' = m(m-1)x^{m-2}\]. These forms will help us simplify and solve the differential equation by substitution.
Simplify the Quadratic Equation
After substituting the derivatives into the original equation, we get: \[x^2 \times m(m-1)x^{m-2} - x \times mx^{m-1} + \frac{10}{9}x^m = 0\]. Simplifying, this becomes:\[m(m-1)x^m - mx^m + \frac{10}{9}x^m = 0\]. To make it easier to solve, we factor out \(x^m\):\[x^m \big[m(m-1) - m + \frac{10}{9}\big] = 0\]. This simplification reduces our problem to solving a quadratic equation in terms of m.
General Solution for Differential Equations
The quadratic equation we derived from the simplification step is: \[m(m-1) - m + \frac{10}{9} = 0\], which simplifies further to: \[m^2 - 2m + \frac{10}{9} = 0\]. Multiplying by 9 to clear the fraction gives:\[9m^2 - 18m + 10 = 0\]. Using the quadratic formula:\[m = \frac{18 \u00b1 \u221a(324 - 360)}{18} = \frac{18 \u00b1 6i}{18} = 1 \u00b1 \frac{1}{3}i\]. So our m values are complex. The general solution is: \[y = x^1 \big[C_1 \text{cos} \big(\frac{1}{3} \text{ln} x\big) + C_2 \text{sin} \big(\frac{1}{3} \text{ln} x\big)\big]\].
Variable Coefficients
This specific equation, \[9x^2 y'' - 9x y' + 10y = 0\], is a second-order differential equation with variable coefficients. This means the coefficients (multiplying the terms) depend on the variable x. Therefore, standard methods like characteristic equations for constant coefficients don鈥檛 work, and other methods like solution form assumptions are necessary.
Differential Equation Solving Steps
Solving second-order linear homogeneous differential equations typically involves key steps:
  • Identify the type of differential equation.
  • Simplify the equation if possible by dividing through with coefficients.
  • Make an assumption for the solution form, e.g., \(y = x^m\).
  • Substitute the assumed form and its derivatives into the equation.
  • Simplify and combine like terms to form a solvable equation.
  • Solve the reduced form, often a quadratic equation, for m.
  • Use the resulting roots to write the general solution.
Following these steps ensures a systematic approach to tackling such differential equations.

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Most popular questions from this chapter

\(x^{3} y^{\prime \prime \prime}+10 x^{2} y^{\prime \prime}-20 x y^{\prime}+20 y=0, y(1)=0\), \(y^{\prime}(1)=-1, y^{\prime \prime}(1)=1\)

\(x^{3} y^{(4)}+6 x^{2} y^{\prime \prime \prime}+7 x y^{\prime \prime}+y^{\prime}=0\)

Suppose the roots of the characteristic equation for the transformed Cauchy- Euler equation are \(r_{1,2}=\alpha \pm \beta i\), so that two solutions are \(y_{1}=x^{\alpha+\beta i}\) and \(y_{2}=x^{\alpha-\beta i}\). Use Euler's formula and the Principle of Superposition to show that \(x^{\alpha} \cos (\beta \ln x)\) and \(x^{\alpha} \sin (\beta \ln x)\) are solutions of the original equation. Hint: \(x^{\beta i}=\left(e^{\ln x}\right)^{\beta i}=\cos (\beta \ln x)+\) \(i \sin (\beta \ln x)\).

Let \(x=\tan t\) so that \(t=\tan ^{-1} x\). (a) Show that \(\frac{d y}{d x}=\frac{1}{1+x^{2}} \frac{d y}{d t}\) and that \(\frac{d^{2} y}{d x^{2}}=\frac{1}{\left(1+x^{2}\right)^{2}}\left(-2 x \frac{d y}{d t}+\frac{d^{2} y}{d t^{2}}\right) .\) (b) Show that this substitution transforms equations of the form $$ \begin{aligned} &a\left(1+x^{2}\right)^{2} y^{\prime \prime}+(2 a x+b)\left(1+x^{2}\right) y^{\prime}+c y \\ &\quad=g(x) \end{aligned} $$ to equations with constant coefficients. (c) Solve \(\left(1+x^{2}\right)^{2} y^{\prime \prime}+2 x\left(1+x^{2}\right) y^{\prime}+4 y=0\). (d) Solve \(\left(1+x^{2}\right)^{2} y^{\prime \prime}+2 x\left(1+x^{2}\right) y^{\prime}+4 y=\) \(\tan ^{-1} x\). (e) Solve \(\left(1+x^{2}\right)^{2} y^{\prime \prime}+2 x\left(1+x^{2}\right) y^{\prime}+4 y=0\), \(y(0)=0, y^{\prime}(0)=1\). (f) Solve \(\left(1+x^{2}\right)^{2} y^{\prime \prime}+2 x\left(1+x^{2}\right) y^{\prime}+4 y=\) \(\tan ^{-1} x, y(0)=0, y^{\prime}(0)=1\).

Show that if \(y_{1}\) is a solution of \(y^{(n)}+\) \(a_{n-1}(t) y^{(n-1)}+\cdots+a_{1}(t) y^{\prime}+a_{0}(t) y=f_{1}(t)\) and \(y_{2}\) is a solution of \(y^{(n)}+a_{n-1}(t) y^{(n-1)}+\) \(\cdots+a_{1}(t) y^{\prime}+a_{0}(t) y=f_{2}(t)\), then \(y_{1}+y_{2}\) is a solution of \(y^{(n)}+a_{n-1}(t) y^{(n-1)}+\cdots+\) \(a_{1}(t) y^{\prime}+a_{0}(t) y=f_{1}(t)+f_{2}(t) .\)
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