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Chapter 4: Problem 1

\(4 x^{2} y^{\prime \prime}-8 x y^{\prime}+5 y=0\)

Short Answer

Expert verified
The general solution is \( y = C_1 x^{2.5} + C_2 x^{0.5} \).

Step by step solution

01

Identify the Type of Differential Equation

The given equation is a second-order linear homogeneous differential equation: \[ 4 x^{2} y'' - 8 x y' + 5 y = 0 \]
02

Assume a Solution of the Form

Consider the solution of the form \[ y = x^r \]. This helps simplify the process of finding the roots and solutions.
03

Differentiate the Assumed Solution

Find the first and second derivatives of \( y = x^r \):\[ y' = r x^{r-1} \]\[ y'' = r (r-1) x^{r-2} \]
04

Substitute the Derivatives into the Original Equation

Substitute \( y = x^r \), \( y' = r x^{r-1} \), and \( y'' = r (r-1) x^{r-2} \) into the original differential equation and simplify:\[ 4 x^{2} r (r-1) x^{r-2} - 8 x r x^{r-1} + 5 x^{r} = 0 \]\[ 4 r (r-1) x^{r} - 8 r x^{r} + 5 x^{r} = 0 \]\[ x^r (4 r (r-1) - 8 r + 5) = 0 \]
05

Solve for the Characteristic Equation

Set the expression inside the parentheses equal to zero to get the characteristic equation:\[ 4 r (r-1) - 8 r + 5 = 0 \]\[ 4 r^2 - 4 r - 8 r + 5 = 0 \]\[ 4 r^2 - 12 r + 5 = 0 \]
06

Solve the Quadratic Equation

Solve the quadratic equation for \( r \):\[ r = \frac{12 \pm \sqrt{(-12)^2 - 4 \, \cdot \, 4 \, \cdot \, 5}}{2 \, \cdot \, 4} \]\[ = \frac{12 \pm \sqrt{144 - 80}}{8} \]\[ = \frac{12 \pm \sqrt{64}}{8} \]\[ = \frac{12 \pm 8}{8} \]\[ = 2.5 \text{ or } 0.5 \]
07

Write the General Solution

Using the roots \( r_1 = 2.5 \) and \( r_2 = 0.5 \), the general solution to the differential equation is:\[ y = C_1 x^{2.5} + C_2 x^{0.5} \]

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Key Concepts

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

Differential Equation
A differential equation is a mathematical equation involving derivatives of a function. In the simplest terms, differential equations describe how a function changes. They are used to model real-world systems, such as motion, heat, and growth.
The equation given in our exercise is a **second-order linear homogeneous differential equation**: \[ 4 x^{2} y'' - 8 x y' + 5 y = 0 \]
This means it has the following characteristics:
  • **Second-order**: The highest derivative is the second derivative (\

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Most popular questions from this chapter

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 \(a d^{2} y / d t^{2}+\) \((b-a) d y / d t+c y=f\left(-e^{t}\right)\).

\(x^{3} y^{\prime \prime \prime}-4 x^{2} y^{\prime \prime}-46 x y^{\prime}+100 y=0\)

\(y^{\prime \prime \prime}+10 y^{\prime \prime}+34 y^{\prime}+40 y=t e^{-4 t}+2 e^{-3 t} \cos t\)

Solve each of the following initial value problems. Verify that your result satisfies the initial conditions by graphing it on an appropriate interval. (a) \(y^{\prime \prime \prime}+3 y^{\prime \prime}+2 y^{\prime}+6 y=0, y(0)=0, y^{\prime}(0)=\) \(1, y^{\prime \prime}(0)=-1\) (b) \(y^{(4)}-8 y^{\prime \prime \prime}+30 y^{\prime \prime}-56 y^{\prime}+49 y=0\), \(y(0)=1, y^{\prime}(0)=2, y^{\prime \prime}(0)=-1, y^{\prime \prime \prime}(0)=\) \(-1\) (c) \(0.31 y^{\prime \prime \prime}+11.2 y^{\prime \prime}-9.8 y^{\prime}+5.3 y=0\), \(y(0)=-1, y^{\prime}(0)=-1, y^{\prime \prime}(0)=0\)

\(9 x^{2} y^{\prime \prime}+27 x y^{\prime}+10 y=x^{-1}, y(1)=0, y^{\prime}(1)=\) \(-1\)
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