Problem 16 Find the general solution. $$y... [FREE SOLUTION]

Chapter 9: Problem 16

Find the general solution. $$y^{\prime \prime}-4 y^{\prime}+4 y=0$$

Short Answer

Expert verified

The general solution for the given differential equation $y^{\prime \prime}-4 y^{\prime}+4 y=0$ is $y(t) = (c_1 + c_2t)e^{2t}$, where $c_1$ and $c_2$ are constants determined by initial conditions.

Step by step solution

Formulate the characteristic equation.

From the given differential equation $y^{\prime \prime}-4 y^{\prime}+4 y=0$, set up the characteristic equation using the coefficients:  r^2 - 4r + 4 = 0 

Solve the characteristic equation.

Next, we need to solve the characteristic equation, which is a quadratic equation. There are several methods to solve quadratic equations, such as factoring or using the quadratic formula. In this case, we can factor the equation easily:  (r - 2)(r - 2) = 0  From this factorization, we can see that the characteristic equation has a double root r = 2.

Write the general solution using the roots.

Since there is a double root, r = 2, we will have to use the second-order linear homogeneous differential equation general solution form for repeated roots:  y(t) = (c_1 + c_2t)e^{rt}  Substituting the value of r in the general solution formula, we get:  y(t) = (c_1 + c_2t)e^{2t} 

Express the final general solution.

The general solution for $y^{\prime \prime}-4 y^{\prime}+4 y=0$ is  y(t) = (c_1 + c_2t)e^{2t}  where c_1 and c_2 are constants determined by initial conditions, if provided.

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Key Concepts

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

Characteristic Equation

The characteristic equation is a key concept when solving linear differential equations, especially homogeneous ones. It provides a bridge between the differential equation and algebraic methods. When given a differential equation like $y'' - 4y' + 4y = 0$, converting it to a characteristic equation helps us find the roots that dictate the solution's behavior.

From the coefficients of the differential equation, form the characteristic equation.
In this example, the characteristic equation is $r^2 - 4r + 4 = 0$.
This is just a quadratic equation involving the variable $r$.

Understanding the characteristic equation is crucial as it simplifies the problem into finding roots, which are much easier to handle than solving differential equations directly.

Quadratic Equation

A quadratic equation is any equation that can be rearranged in the form $ax^2 + bx + c = 0$. In the context of differential equations, it represents the characteristic equation. It can be solved using several methods:

Factoring, which is simple when the equation can be easily split into two binomial expressions.
The quadratic formula: $r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, useful for more complex cases.
Completing the square, an alternative method.

In our problem, the quadratic $r^2 - 4r + 4$ factored neatly into $(r-2)(r-2) = 0$. This provided a repeated root, showing that $r = 2$ appears twice, indicating a unique type of solution in the realm of differential equations.

Repeated Roots

Repeated roots occur in quadratic and other polynomial characteristic equations when a root appears more than once, as can happen when solving differential equations.

In the characteristic equation $r^2 - 4r + 4 = 0$, the root $r = 2$ is repeated because the equation factors into $(r-2)^2 = 0$.
Repeated roots suggest that the usual solutions need adjustment, often introducing polynomial multiplication to reflect the multiplicity.
The general solution takes the form $y(t) = (c_1 + c_2t)e^{rt}$, incorporating a term involving $t$ to account for the root's repetition.

This adjusted solution ensures the solution's completeness, addressing the unique structure repeated roots introduce.

Homogeneous Differential Equations

A homogeneous differential equation is one where the function and its derivatives are set equal to zero, traditionally written in the form $a_n y^{(n)} + a_{n-1} y^{(n-1)} + \ldots + a_1 y' + a_0 y = 0$. Solving these involves seeking solutions of the form $y(t) = e^{rt}$:

Finding the characteristic equation is the first step, which involves transforming the original differential equation.
Solutions are dictated by the roots of the characteristic equation. These roots can be real and distinct, real and repeated, or complex.
For repeated roots, such as with $y'' - 4y' + 4y = 0$, the solution incorporates terms to reflect this multiplicity.

Thus, understanding homogeneous differential equations and how to solve them through their characteristic equations allows one to solve a broad class of problems efficiently.

91影视

Find the general solution. $$y^{\prime \prime}-4 y^{\prime}+4 y=0$$

Short Answer

Step by step solution

Formulate the characteristic equation.

Solve the characteristic equation.

Write the general solution using the roots.

Express the final general solution.

Key Concepts

Characteristic Equation

Quadratic Equation

Repeated Roots

Homogeneous Differential Equations

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