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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.
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:
  • \[r^4 + 6r^3 = 0\]
Solving this equation helps us determine the nature of the solutions, which can be real, repeated, or complex roots.
In this particular problem, the characteristic equation's factorization gives us the roots:
  • \[r_1 = 0\]
  • \[r_2 = 0\]
  • \[r_3 = 0\]
  • \[r_4 = -6\]
These roots are then used to form the general solution for the function \(u(t)\). In this way, the characteristic equation is crucial for simplifying the original differential equation into an algebraic form that is easier to solve.
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
  • \[t^{2}(t+3)y^{\prime \prime \prime}-3t(t+2)y^{\prime \prime}+6(1+t)y^{\prime}-6y=0\]
is homogeneous because all terms involving the function and its derivatives equal zero. This property means we are looking for solutions that satisfy the condition without any external forcing function or non-homogeneous term.
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.

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Most popular questions from this chapter

Determine the general solution of the given differential equation. \(y^{\prime \prime \prime}-y^{\prime}=2 \sin t\)

Find the general solution of the given differential equation. $$ y^{\mathrm{iv}}+6 y^{\prime \prime \prime}+17 y^{\prime \prime}+22 y^{\prime}+14 y=0 $$

determine intervals in which solutions are sure to exist. $$ \left(x^{2}-4\right) y^{\mathrm{vi}}+x^{2} y^{\prime \prime \prime}+9 y=0 $$

Express the given complex number in the form \(R(\cos \theta+\) \(i \sin \theta)=R e^{i \theta}\) $$ 1+i $$

We consider another way of arriving at the proper form of \(Y(t)\) for use in the method of undetermined coefficients. The procedure is based on the observation that exponential, polynomial, or sinusoidal terms (or sums and products of such terms) can be viewed as solutions of certain linear homogeneous differential equations with constant coefficients. It is convenient to use the symbol \(D\) for \(d / d t\). Then, for example, \(e^{-t}\) is a solution of \((D+1) y=0 ;\) the differential operator \(D+1\) is said to annihilate, or to be an annihilator of, \(e^{-t}\). Similarly, \(D^{2}+4\) is an annihilator of \(\sin 2 t\) or \(\cos 2 t,\) \((D-3)^{2}=D^{2}-6 D+9\) is an annihilator of \(e^{3 t}\) or \(t e^{3 t},\) and so forth. Consider the problem of finding the form of the particular solution \(Y(t)\) of $$ (D-2)^{3}(D+1) Y=3 e^{2 t}-t e^{-t} $$ where the left side of the equation is written in a form corresponding to the factorization of the characteristic polynomial. (a) Show that \(D-2\) and \((D+1)^{2},\) respectively, are annihilators of the terms on the right side of Eq. and that the combined operator \((D-2)(D+1)^{2}\) annihilates both terms on the right side of Eq. (i) simultaneously. (b) Apply the operator \((D-2)(D+1)^{2}\) to Eq. (i) and use the result of Problem 20 to obtain $$ (D-2)^{4}(D+1)^{3} Y=0 $$ Thus \(Y\) is a solution of the homogeneous equation (ii). By solving Eq. (ii), show that $$ \begin{aligned} Y(t)=c_{1} e^{2 t} &+c_{2} t e^{2 t}+c_{3} t^{2} e^{2 t}+c_{4} t^{3} e^{2 t} \\ &+c_{5} e^{-t}+c_{6} t e^{-t}+c_{7} t^{2} e^{-t} \end{aligned} $$ where \(c_{1}, \ldots, c_{7}\) are constants, as yet undetermined. (c) Observe that \(e^{2 t}, t e^{2 t}, t^{2} e^{2 t},\) and \(e^{-t}\) are solutions of the homogeneous equation corresponding to Eq. (i); hence these terms are not useful in solving the nonhomogeneous equation. Therefore, choose \(c_{1}, c_{2}, c_{3},\) and \(c_{5}\) to be zero in Eq. (iii), so that $$ Y(t)=c_{4} t^{3} e^{2 t}+c_{6} t e^{-t}+c_{7} t^{2} e^{-t} $$ This is the form of the particular solution \(Y\) of Eq. (i). The values of the coefficients \(c_{4}, c_{6},\) and \(c_{7}\) can be found by substituting from Eq. (iv) in the differential equation (i).

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