Chapter 4: Problem 28
Use the method of reduction of order (Problem 26) to solve the given differential equation. $$ t^{2}(t+3) y^{\prime \prime \prime}-3 t(t+2) y^{\prime \prime}+6(1+t) y^{\prime}-6 y=0, \quad t>0 ; \quad y_{1}(t)=t^{2}, \quad y_{2}(t)=t^{3} $$
Short Answer
Expert verified
Answer: The third solution is given by \(y_{3}(t) = c_1t^2 + c_2t^3 + c_3t^4 + c_4t^2e^{-6t} - 6c_4t^3e^{-6t}\), where \(c_1\), \(c_2\), \(c_3\), and \(c_4\) are constants.
Step by step solution
01
Introducing a new function
Let's introduce a new function: \(u(t)\). Then we can represent the third solution as \(y_{3}(t) = u(t)y_{1}(t) + u'(t)y_{2}(t)\). Our goal will be to find this function u(t) and then obtain the third solution.
02
Calculate the first and second derivatives of the third solution
We will write the first and second derivatives of the third solution \(y_{3}(t)\) using the function u(t) and its derivatives.
First derivative: \(y_{3}'(t) = u'(t)y_{1}(t) + u(t)y_{1}'(t) + u''(t)y_{2}(t) + u'(t)y_{2}'(t)\)
Second derivative: \(y_{3}''(t) = u''(t)y_{1}(t) + 2u'(t)y_{1}'(t) + u(t)y_{1}''(t) + u'''(t)y_{2}(t) + 2u''(t)y_{2}'(t) + u'(t)y_{2}''(t)\)
Now we must find the third derivative of \(y_3(t)\).
Third derivative: \(y_{3}'''(t) = u'''(t)y_{1}(t) + 3u''(t)y_{1}'(t) + 3u'(t)y_{1}''(t) + u(t)y_{1}'''(t) + u''''(t)y_{2}(t) + 3u'''(t)y_{2}'(t) + 3u''(t)y_{2}''(t) + u'(t)y_{2}'''(t)\)
03
Substitute the given values into the equation
Now, we will substitute these expressions for \(y_{3}(t)\) and its derivatives into the given differential equation. Keep in mind that we know the following:
\(y_{1}(t) = t^2\)
\(y_{2}(t) = t^3\)
Therefore,
\(y_{1}'(t) = 2t\)
\(y_{2}'(t) = 3t^2\)
\(y_{1}''(t) = 2\)
\(y_{2}''(t) = 6t\)
\(y_{1}'''(t) = 0\)
\(y_{2}'''(t) = 6\)
Now, we can substitute all these values and the expressions for the derivatives of \(y_3(t)\) into the given differential equation. After simplifying the equation, we obtain the following equation:
\(u''''(t) + 6u'''(t) = 0\)
04
Solve the differential equation for u(t)
Now we have a homogeneous linear differential equation with constant coefficients for u(t). We can find the general solution for u(t) using standard techniques. We first obtain the characteristic equation:
\(r^4 + 6r^3 = 0\)
\(r^3(r+6) = 0\)
So, the roots are \(r_1 = 0\), \(r_2 = 0\), \(r_3 = 0\), \(r_4 = -6\). Thus, the general solution for u(t) is:
\(u(t) = c_1 + c_2t + c_3t^2 + c_4e^{-6t}\)
05
Find the third solution y3(t)
Now that we have the function u(t), we can find the third solution \(y_3(t)\):
\(y_{3}(t) = u(t)y_{1}(t) + u'(t)y_{2}(t) = (c_1 + c_2t + c_3t^2 + c_4e^{-6t})t^2 + (c_2 + 2c_3t - 6c_4e^{-6t})t^3\)
And so, the third solution is given by:
\(y_{3}(t) = c_1t^2 + c_2t^3 + c_3t^4 + c_4t^2e^{-6t} - 6c_4t^3e^{-6t}\)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Differential Equations
Differential equations are mathematical equations that involve functions and their derivatives. They are used to describe phenomena where there are dependent changes, such as in physics, engineering, and other sciences.
Broadly speaking, these equations tell us how a particular variable changes over time or space depending on other variables and their rates of change. In the given exercise, we deal with a third-order differential equation. This means it involves the third derivative of the function. This type of equation can often be more complex to solve compared to first-order or second-order differential equations, due to the higher level of derivatives involved.
The process of solving differential equations typically involves finding the function or functions that satisfy the given equation. In this exercise, the method of reduction of order helps in reducing the complexity by breaking down the problem, making it easier to find the solutions. This method is particularly useful when we already have known solutions to the differential equation, and we use them to discover new solutions.
Broadly speaking, these equations tell us how a particular variable changes over time or space depending on other variables and their rates of change. In the given exercise, we deal with a third-order differential equation. This means it involves the third derivative of the function. This type of equation can often be more complex to solve compared to first-order or second-order differential equations, due to the higher level of derivatives involved.
The process of solving differential equations typically involves finding the function or functions that satisfy the given equation. In this exercise, the method of reduction of order helps in reducing the complexity by breaking down the problem, making it easier to find the solutions. This method is particularly useful when we already have known solutions to the differential equation, and we use them to discover new solutions.
Characteristic Equation
A characteristic equation is an algebraic equation derived from a differential equation and is used to find the roots, or solutions, to the differential equation.
It arises from the substitution of a trial solution into a linear differential equation with constant coefficients. For the equation in the exercise, we find that the characteristic equation is given by:
In this particular problem, the characteristic equation's factorization gives us the roots:
It arises from the substitution of a trial solution into a linear differential equation with constant coefficients. For the equation in the exercise, we find that the characteristic equation is given by:
- \[r^4 + 6r^3 = 0\]
In this particular problem, the characteristic equation's factorization gives us the roots:
- \[r_1 = 0\]
- \[r_2 = 0\]
- \[r_3 = 0\]
- \[r_4 = -6\]
Homogeneous Linear Equations
Homogeneous linear equations are a specific type of linear differential equation where the function and its derivatives sum to zero. The concept of homogeneity implies that the equation is balanced out by solutions of certain forms that include functions of each other and their derivatives.
In the given exercise, the equation
Solving such equations often involves techniques like the method of reduction of order, as used in this problem, allowing for a more manageable way to find particular and general solutions when certain solutions are already known. Understanding and identifying homogeneous equations are important because they provide insights into the natural behavior of systems modeled by such equations without external influence.
In the given exercise, the equation
- \[t^{2}(t+3)y^{\prime \prime \prime}-3t(t+2)y^{\prime \prime}+6(1+t)y^{\prime}-6y=0\]
Solving such equations often involves techniques like the method of reduction of order, as used in this problem, allowing for a more manageable way to find particular and general solutions when certain solutions are already known. Understanding and identifying homogeneous equations are important because they provide insights into the natural behavior of systems modeled by such equations without external influence.