Logout Open In App
Open In App
Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Chapter 7: Problem 63

In Exercises \(61-68\) use the method suggested by Exercise 60 to find the general solution on some interval \((0, \rho)\) $$ 28 x^{2}(1-3 x) y^{\prime \prime}-7 x(5+9 x) y^{\prime}+7(2+9 x) y=0 $$

Short Answer

Expert verified
The general solution of the given differential equation within the interval (0, ρ) is: $$ y(x) = C_1(x^2(1-3x))^{r_1} + C_2(x^2(1-3x))^{r_2} $$ where \(C_1\) and \(C_2\) are constants, and \(r_1\) and \(r_2\) are found by solving the quadratic equation obtained in Step 4.

Step by step solution

01

Introduce the variable substitution

Let \(u = x^2(1 - 3x)\). The given equation becomes: $$ 28u\frac{d^2y}{du^2} - 7(5+9u)\frac{dy}{du} +7(2+9u)y = 0 $$ Now, we will try to find a second substitution to turn this into an equation with constant coefficients.
02

Introduce the second variable substitution

Let \(z(u) = y(u)\). Now, we can rewrite the equation in terms of \(z\): $$ 28u\frac{d^2z}{du^2} - 7(5+9u)\frac{dz}{du} +7(2+9u)z = 0 $$ It is a Cauchy-Euler type equation, which has the general form: $$ au^2\frac{d^2z}{du^2} + b\ \!u\frac{dz}{du} + c\ \!z = 0 $$ with \(a=28\), \(b=-7(5+9u)\), and \(c=7(2+9u)\).
03

Solve the Cauchy-Euler equation

To solve this type of equation, we try the solution of the form \(z(u) = u^r\): $$ 28u^2r(r-1)u^{r-2} - 7(5+9u)ru^{r-1} +7(2+9u)u^r = 0 $$ Multiplying through by \(u^{2-r}\), we obtain: $$ 28r(r-1) - 7(5+9u)r +7(2+9u)u^2 = 0 $$ This is now a quadratic equation in r: $$ 28r^2 - 35r - 9ru + 63u^2 = 0 $$
04

Solve for r

The discriminant (Δ) of the quadratic equation is: $$ Δ = (35 - 9u)^2 - 4\cdot 28\cdot 63u^2 $$ Upon factoring and solving for r, we obtain: $$ r_{1,2} = \frac{35\pm\sqrt{Δ}}{56} $$ However, since we are looking for a solution within an interval \((0, \rho)\), we will choose the values of r that satisfy this condition. From here, we will now write down the general solution.
05

Write down the general solution

The general solution for the Cauchy-Euler equation is given as the sum of independent solutions with the calculated values of r: $$ z(u) = C_1u^{r_1} + C_2u^{r_2} $$ Now, we revert the variable substitution back to the original problem, that is: $$ y(x) = z(x^2(1-3x)) = C_1(x^2(1-3x))^{r_1} + C_2(x^2(1-3x))^{r_2} $$ This is the general solution of the given differential equation within the interval \((0, \rho)\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Variable Substitution
In the realm of differential equations, variable substitution is a powerful technique used to simplify complex equations. The key idea is to introduce new variables that transform the original problem into a more manageable form. In our exercise, we start by redefining the variable using the substitution \(u = x^2(1 - 3x)\). This step helps to translate the initial complex equation into a more straightforward one in terms of \(u\). This rewritten equation allows us to utilize methods dedicated to solving equations with certain characteristics, such as the Cauchy-Euler form.
Using the variable substitution not only makes the problem more tractable but also opens the door to applying standard solution techniques that might not be obvious in the initial form. By doing so, we can uncover the underlying structure of the differential equation, making the solution process more systematic. Variable substitutions are crucial in transforming and solving various forms of differential equations, leading us into steps where methods like the Cauchy-Euler approach become applicable.
Quadratic Equation
Quadratic equations frequently appear when solving Cauchy-Euler differential equations. After substituting the variables, we often end up with a polynomial equation related to the order of the differential equation. This polynomial is typically quadratic in our step-by-step solution.
  • The quadratic equation mentioned \(28r^2 - 35r - 9ru + 63u^2 = 0\) arises from assuming solutions of the form \(z(u) = u^r\).
  • Here, \(r\) represents the roots which define the characteristic behavior of the solution.
Quadratic equations, such as this one, can be solved through finding roots using the quadratic formula. The roots \(r_1\) and \(r_2\) derived from this equation are crucial as they determine the exponent values in our general solution, reflecting the nature of the solutions across an interval.
This step emphasizes the importance of understanding how to analyze and solve quadratic equations, as these directly contribute to constructing the complete solution of the differential equation.
Differential Equation Solution
Solving differential equations involves creating a function that satisfies the given equation. The original problem involves a form of the Cauchy-Euler equation. Such equations are characterized by their coefficients being powers of the independent variable. These equations often demand a solution approach different from standard differential equations.
The typical strategy involves seeking solutions structured as \(z(u) = u^r\), where \(r\) is determined from the quadratic auxiliary equation we discussed earlier. Solving this leads us to find a general solution composed of terms related to these \(r\) values:
  • Each root, \(r_1\) and \(r_2\), contributes to the solution in the form \(C_1u^{r_1} + C_2u^{r_2}\).
This solution must be subsequently converted back to the terms of the original variable, using the initial substitution \(u = x^2(1-3x)\), resulting in the form:
\[y(x) = C_1(x^2(1-3x))^{r_1} + C_2(x^2(1-3x))^{r_2}\]
This final expression represents the general solution of the original differential equation, satisfying the problem in context.
Interval of Validity
The interval of validity is a critical part when finding the solution to a differential equation. It defines where the solution is applicable and ensures the results are consistent with the initial problem conditions. In the step-by-step solution, the interval \(0, \rho)\) is discussed.
This interval is chosen based on the behavior of the original differential equation and any restrictions that occur from transformations or singularities found in substitutions. It is essentially a range within which the solution \(y(x)\) is valid and behaves according to the model provided by the differential equation.
When applying the substitutions and finding the roots \(r_1\) and \(r_2\), it is essential to check if the assumptions and transformations made are valid within this interval and that no undefined operations, such as division by zero, occur. Evaluating the interval ensures that the solution is more than just a mathematical exercise; it confirms the solution holds true under the problem's physical or theoretical constraints.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

In Exercises \(61-68\) use the method suggested by Exercise 60 to find the general solution on some interval \((0, \rho)\) $$ 4 x^{2}\left(3+2 x+x^{2}\right) y^{\prime \prime}-x\left(3-14 x-15 x^{2}\right) y^{\prime}+\left(3+7 x^{2}\right) y=0 $$

Find power series in \(x\) for the solutions \(y_{1}\) and \(y_{2}\) of $$ y^{\prime \prime}+4 x y^{\prime}+\left(2+4 x^{2}\right) y=0 $$ such that \(y_{1}(0)=1, y_{1}^{\prime}(0)=0, y_{2}(0)=0, y_{2}^{\prime}(0)=1,\) and identify \(y_{1}\) and \(y_{2}\) in terms of familiar elementary functions.

Find \(a_{0}, \ldots, a_{N}\) for \(N\) at least 7 in the power series \(y=\sum_{n=0}^{\infty} a_{n}\left(x-x_{0}\right)^{n}\) for the solution of the initial value problem. Take \(x_{0}\) to be the point where the initial conditions are imposed. $$ \left(x^{2}-4\right) y^{\prime \prime}-x y^{\prime}-3 y=0, \quad y(0)=-1, \quad y^{\prime}(0)=2 $$

Let $$L y=x^{2}\left(\alpha_{0}+\alpha_{q} x^{q}\right) y^{\prime \prime}+x\left(\beta_{0}+\beta_{q} x^{q}\right) y^{\prime}+\left(\gamma_{0}+\gamma_{q} x^{q}\right) y$$ where \(q\) is a positive integer, and define $$p_{0}(r)=\alpha_{0} r(r-1)+\beta_{0} r+\gamma_{0} \quad \text { and } \quad p_{q}(r)=\alpha_{q} r(r-1)+\beta_{q} r+\gamma_{q}$$ Suppose $$p_{0}(r)=\alpha_{0}\left(r-r_{1}\right)^{2} \quad \text { and } \quad p_{q}(r) \not \equiv 0$$ (a) Recall from Exercise 7.5 .59 that \(L y=0\) has the solution $$y_{1}=x^{r_{1}} \sum_{m=0}^{\infty} a_{q m}\left(r_{1}\right) x^{q m}$$ where $$a_{q m}\left(r_{1}\right)=\frac{(-1)^{m}}{\left(q^{2} \alpha_{0}\right)^{m}(m !)^{2}} \prod_{j=1}^{m} p_{q}\left(q(j-1)+r_{1}\right)$$ (b) Show that \(L y=0\) has the second solution $$y_{2}=y_{1} \ln x+x^{r_{1}} \sum_{m=1}^{\infty} a_{q m}^{\prime}\left(r_{1}\right) J_{m} x^{q m}$$ where $$J_{m}=\sum_{j=1}^{m} \frac{p_{q}^{\prime}\left(q(j-1)+r_{1}\right)}{p_{q}\left(q(j-1)+r_{1}\right)}-\frac{2}{q} \sum_{j=1}^{m} \frac{1}{j} .$$ (c) Conclude from (a) and (b) that if \(\gamma_{q} \neq 0\) then $$y_{1}=x^{r_{1}} \sum_{m=0}^{\infty} \frac{(-1)^{m}}{(m !)^{2}}\left(\frac{\gamma_{q}}{q^{2} \alpha_{0}}\right)^{m} x^{q m}$$ Let $$L y=x^{2}\left(\alpha_{0}+\alpha_{q} x^{q}\right) y^{\prime \prime}+x\left(\beta_{0}+\beta_{q} x^{q}\right) y^{\prime}+\left(\gamma_{0}+\gamma_{q} x^{q}\right) y$$ where \(q\) is a positive integer, and define $$p_{0}(r)=\alpha_{0} r(r-1)+\beta_{0} r+\gamma_{0} \quad \text { and } \quad p_{q}(r)=\alpha_{q} r(r-1)+\beta_{q} r+\gamma_{q}$$ Suppose $$p_{0}(r)=\alpha_{0}\left(r-r_{1}\right)^{2} \quad \text { and } \quad p_{q}(r) \not \equiv 0$$ (a) Recall from Exercise 7.5 .59 that \(L y=0\) has the solution $$y_{1}=x^{r_{1}} \sum_{m=0}^{\infty} a_{q m}\left(r_{1}\right) x^{q m}$$ where $$a_{q m}\left(r_{1}\right)=\frac{(-1)^{m}}{\left(q^{2} \alpha_{0}\right)^{m}(m !)^{2}} \prod_{j=1}^{m} p_{q}\left(q(j-1)+r_{1}\right)$$ (b) Show that \(L y=0\) has the second solution $$y_{2}=y_{1} \ln x+x^{r_{1}} \sum_{m=1}^{\infty} a_{q m}^{\prime}\left(r_{1}\right) J_{m} x^{q m}$$ where $$J_{m}=\sum_{j=1}^{m} \frac{p_{q}^{\prime}\left(q(j-1)+r_{1}\right)}{p_{q}\left(q(j-1)+r_{1}\right)}-\frac{2}{q} \sum_{j=1}^{m} \frac{1}{j}$$ (c) Conclude from (a) and (b) that if \(\gamma_{q} \neq 0\) then $$y_{1}=x^{r_{1}} \sum_{m=0}^{\infty} \frac{(-1)^{m}}{(m !)^{2}}\left(\frac{\gamma_{q}}{q^{2} \alpha_{0}}\right)^{m} x^{q m}$$ are solutions of $$ \alpha_{0} x^{2} y^{\prime \prime}+\beta_{0} x y^{\prime}+\left(\gamma_{0}+\gamma_{q} x^{q}\right) y=0. $$

Find a fundamental set of Frobenius solutions. Give explicit formulas for the coefficients. \(3 x^{2}\left(3+x^{2}\right) y^{\prime \prime}+x\left(3+11 x^{2}\right) y^{\prime}+\left(1+5 x^{2}\right) y=0\)
See all solutions

Recommended explanations on Math Textbooks

Calculus

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Decision Maths

Read Explanation

Geometry

Read Explanation

Statistics

Read Explanation

Logic and Functions

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free