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Solve the given differential equation. $$ x^{2} y^{\prime \prime}-2 y=0 $$

Short Answer

Expert verified
The solution is \( y(x) = C_1 x^2 + C_2 x^{-1} \), with constants \( C_1 \) and \( C_2 \).

Step by step solution

01

Identify the Type of Differential Equation

The given differential equation is \( x^2y'' - 2y = 0 \). It is a second-order linear homogeneous differential equation with variable coefficients.
02

Look for a Suitable Substitution

For equations of this form, consider a solution of the type \( y = x^m \), where \( m \) is a constant that we need to determine.
03

Substitute and Differentiate

Substitute \( y = x^m \) into the equation and find the derivatives. We have \( y' = mx^{m-1} \) and \( y'' = m(m-1)x^{m-2} \).
04

Plug Derivatives into the Original Equation

Substituting these into the original differential equation, we get: \[ x^2 (m(m-1)x^{m-2}) - 2x^m = 0. \] Simplify this expression: \[ m(m-1)x^m - 2x^m = 0. \]
05

Factor and Solve for m

Factor out \( x^m \): \[ x^m (m(m-1) - 2) = 0. \] Since \( x^m \) is never zero, solve for \( m \) in \( m(m-1) - 2 = 0 \).
06

Solve the Quadratic Equation

The equation is \( m^2 - m - 2 = 0 \). Solve this using the quadratic formula \( m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = -1 \), and \( c = -2 \).
07

Find the Roots

Calculate the discriminant: \( (-1)^2 - 4(1)(-2) = 1 + 8 = 9 \). Therefore, \( m = \frac{1 \pm 3}{2} \), giving \( m = 2 \) and \( m = -1 \).
08

Write the General Solution

With roots \( m = 2 \) and \( m = -1 \), the general solution is a linear combination: \[ y(x) = C_1x^2 + C_2x^{-1}, \] where \( C_1 \) and \( C_2 \) are constants.

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Key Concepts

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

Second-order Linear Homogeneous Differential Equation
Differential equations are mathematical equations that involve functions and their derivatives. Among them, second-order linear homogeneous differential equations are crucial in many fields, including physics and engineering. This type of differential equation has the form:
  • The highest derivative is of order two, implying it's a second-order equation.
  • The term 'homogeneous' indicates that every part of the equation contains the function or its derivatives (e.g., there is no standalone constant term).
  • It's linear if each term is either a constant, the dependent variable, or any of its derivatives with no products.

For example, the equation:
\[ x^2y'' - 2y = 0 \]
is of this type. Understanding these equations is key to solving more complex problems involving dynamic systems.By finding a solution form like \( y = x^m \), we transform the problem into a simpler algebraic form, facilitating the solution process.
Variable Coefficients
In some differential equations, the coefficients are not constants but functions of the independent variable. Such equations are known as differential equations with variable coefficients.
This characteristic adds a layer of complexity to the problem. A good example is the term \( x^2 \) in the differential equation \( x^2y'' - 2y = 0 \).
Here, the coefficient \( x^2 \) is not a constant but varies with \( x \).

Why does this matter? Variable coefficients mean that solutions cannot typically be expressed in simple exponents or trigonometric functions. Often, special techniques or methods, like power series or substitution methods (e.g., assuming \( y = x^m \)), are used to find solutions.
This approach transforms the differential equation into an algebraic equation, allowing us to isolate terms and solve for unknowns effectively.
Quadratic Formula
The quadratic formula is a powerful tool used to solve quadratic equations of the form \( ax^2 + bx + c = 0 \). The solution is given by:
  • \( m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

In the context of differential equations, especially when dealing with variable powers like \( y = x^m \), the quadratic formula helps in determining the roots related to the characteristic equation derived from substituting assumed solutions.
For instance, solving the equation \( m^2 - m - 2 = 0 \) involves this formula:
  • Here, \( a = 1 \), \( b = -1 \), \( c = -2 \).
  • The roots \( m = 2 \) and \( m = -1 \) are found by applying the formula.
Once roots are found, they lead directly to the general solution for the differential equation, such as \( y(x) = C_1x^2 + C_2x^{-1} \). Knowing how to apply this formula is essential in solving second-order equations with variable coefficients.

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Most popular questions from this chapter

Consider a pendulum that is released from rest from an initial displacement of \(\theta_{0}\) radians. Solving the linear model (7) subject to the initial conditions \(\theta(0)=\theta_{0}, \theta^{\prime}(0)=0\) gives \(\theta(t)=\theta_{0} \cos \sqrt{g} \| t .\) Theperiod of oscillations predicted by this modelisgivenbythefamiliarformula \(T=2 \pi / \sqrt{g l l}=2 \pi \sqrt{U g}\). The interesting thing about this formula for \(T\) is that it does not depend on the magnitude of the initial displacement \(\theta_{0}\). In other words, the linear model predicts that the time that it would take the pendulum to swing from an initial displacement of, say, \(\theta_{0}=\pi / 2\left(=90^{\circ}\right)\) to \(-\pi / 2\) and back again would be exactly the same time to cycle from, say, \(\theta_{0}=\pi / 360\left(=0.5^{\circ}\right)\) to \(-\pi / 360\). This is intuitively unreasonable; the actual period must depend on \(\theta_{0}\). If we assume that \(g=32 \mathrm{ft} / \mathrm{s}^{2}\) and \(l=32 \mathrm{ft}\), then the period of oscillation of the linear model is \(T=2 \pi \mathrm{s}\). Let us compare this last number with the period predicted by the nonlinear model when \(\theta_{0}=\pi / 4\). Using a numerical solver that is capable of generating hard data, approximate the solution of $$\frac{d^{2} \theta}{d t^{2}}+\sin \theta=0, \quad \theta(0)=\frac{\pi}{4}, \quad \theta^{\prime}(0)=0$$ for \(0 \leq t \leq 2\). As in Problem 24 , if \(t_{1}\) denotes the first time the pendulum reaches the position \(O P\) in Figure 3.11.3, then the period of the nonlinear pendulum is \(4 t_{1} .\) Here is another way of solving the equation \(\theta(t)=0\). Expeniment with small step sizes and advance the time staning at \(t=0\) and ending at \(t=2\). From your hard data, observe the time \(t_{1}\) when \(\theta(t)\) changes, for the first time, from positive to negative. Use the value \(t_{1}\) to determine the true value of the period of the nonlinear pendulum. Compute the percentage relative error in the period estimated by \(T=2 \pi\).

Solve the given system of differential equations by systematic elimination. $$ \begin{aligned} &2 \frac{d x}{d t}-5 x+\frac{d y}{d t}=e^{t} \\ &\frac{d x}{d t}-x+\frac{d y}{d t}=5 e^{t} \end{aligned} $$

Find the general solution of \(x^{4} y^{\prime \prime}+x^{3} y^{\prime}-4 x^{2} y=1\) given that \(y_{1}=x^{2}\) is a solution of the associated homogeneous equation.

Find the charge on the capacitor in an \(L R C\) -series circuit when \(L=\frac{1}{2} \mathrm{~h}, R=10 \Omega, C=0.01 \mathrm{f}, E(t)=150 \mathrm{~V}, q(0)=1 \mathrm{C}\), and \(i(0)=0 \mathrm{~A} .\) What is the charge on the capacitor after a long time?

Solve the given differential equation by undetermined coefficients. \(y^{\prime \prime}+2 y^{\prime}+y=\sin x+3 \cos 2 x\)

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