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Chapter 3: Problem 18

Show that \(y=x^{3}\) is a solution of \(y y^{\prime \prime}=6 x^{4}\), but that if \(c^{2} \neq 1\), then \(y=c x^{3}\) is not a solution.

Short Answer

Expert verified
Yes, \( y = x^3 \) is a solution, but \( y = cx^3 \) is not if \( c^2 \neq 1 \).

Step by step solution

01

Identify the Given Functions and Equations

We need to show that the function \( y = x^3 \) satisfies the equation \( y y'' = 6x^4 \), and if \( y = cx^3 \) for some constant \( c^2 eq 1 \), it does not satisfy this equation.
02

Differentiate the Function y = x^3

Since \( y = x^3 \), find the second derivative. The first derivative \( y' = \frac{d}{dx}(x^3) = 3x^2 \). The second derivative \( y'' = \frac{d^2}{dx^2}(x^3) = 6x \).
03

Verify Solution for y = x^3

Substitute \( y = x^3 \) and \( y'' = 6x \) into the equation \( y y'' = 6x^4 \). We have \( x^3 \cdot 6x = 6x^4 \). Simplifying gives \( 6x^4 = 6x^4 \), confirming \( y = x^3 \) is a solution.
04

Trial Function y = cx^3

Assume \( y = cx^3 \) and find its second derivative. First derivative \( y' = 3cx^2 \), and second derivative \( y'' = 6cx \).
05

Check Condition for y = cx^3

Substitute \( y = cx^3 \) and \( y'' = 6cx \) into the equation \( y y'' = 6x^4 \). We get \( cx^3 \cdot 6cx = 6c^2x^4 \).
06

Analyze the Result and Consider Parameter Conditions

Equation becomes \( 6c^2x^4 = 6x^4 \). For equality, \( c^2 \) must equal 1. If \( c^2 eq 1 \), then \( 6c^2x^4 eq 6x^4 \). Hence \( y = cx^3 \) is not a solution if \( c^2 eq 1 \).

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Key Concepts

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

Second Derivative
Differentiation helps us understand how a function changes at any given point. In the context of differential equations, the second derivative is particularly important because it tells us about the curvature of the function.
For a given function like \( y = x^3 \), the process of finding derivatives follows clear rules:
  • First derivative \( y' = \frac{d}{dx}(x^3) = 3x^2 \) captures the rate of change or slope of the function.
  • Second derivative \( y'' = \frac{d^2}{dx^2}(x^3) = 6x \) provides information about the rate at which this slope changes, essentially how much the curve bends.
Understanding the second derivative is crucial for solving differential equations and verifying whether a given function satisfies them. It allows us to link the changes in functions with real-world processes that might also change over time, like motion or growth patterns.
Trial Function
A trial function is a proposed solution to a differential equation. It acts as a guess that we verify through substitution into the differential equation.
For instance, suppose we propose \( y = cx^3 \) as a potential solution to the equation \( y y'' = 6x^4 \). Here, \( c \) is a constant whose value we must determine.
  • The trial function \( y = cx^3 \) gives the first derivative \( y' = 3cx^2 \) and the second derivative \( y'' = 6cx \).
  • Substituting these into the differential equation helps us check if the proposal holds for all values of \( x \).
The idea behind using a trial function is to see if our guess satisfies the equation under a given set of constraints. It's a starting point that, with further verification, might reveal a genuine solution or show what's wrong with the initial guess.
Verifying Solutions
Once a function is proposed as a solution to a differential equation, verifying its correctness is key. This involves plugging the function and its derivatives back into the equation.
For example:
  • Initially, check \( y = x^3 \) and its second derivative \( y'' = 6x \). Substituting these into \( y y'' = 6x^4 \) results in the true statement \( 6x^4 = 6x^4 \), confirming it as a solution.
  • For \( y = cx^3 \), after substitution, the equation modifies to \( 6c^2x^4 = 6x^4 \). Here, \( c^2 \) must be 1 to hold this equality. Any other value of \( c \) disproves it as a solution.
Verifying solutions is a critical step because not all proposed solutions will satisfy the differential equation. By working through substitution and simplification, we confirm whether our function works in every possible scenario defined by the original equation.

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Most popular questions from this chapter

Find a particular solution \(y_{p}\) of the given equation. In all these problems, primes denote derivatives with respect to \(x\). $$ y^{(4)}-5 y^{\prime \prime}+4 y=e^{x}-x e^{2 x} $$

Find a particular solution \(y_{p}\) of the given equation. In all these problems, primes denote derivatives with respect to \(x\). $$ y^{(3)}-y=e^{x}+7 $$

Set up the appropriate form of a particular solution \(y_{p}\), but do not determine the values of the coefficients. $$ y^{\prime \prime}-2 y^{\prime}+2 y=e^{x} \sin x $$

Problems pertain to the solution of differential equations with complex coefficients. (a) Use Euler's formula to show that every complex number can be written in the form \(r e^{i \theta}\), where \(r \geqq 0\) and \(-\pi<\theta \leqq \pi\) (b) Express the numbers \(4,-2,3 i\) \(1+i\), and \(-1+i \sqrt{3}\) in the form \(r e^{i \theta}\). (c) The two square roots of \(r e^{i \theta}\) are \(\pm \sqrt{r} e^{i \theta / 2} .\) Find the square roots of the numbers \(2-2 i \sqrt{3}\) and \(-2+2 i \sqrt{3}\).

Verify that \(y_{1}=x\) and \(y_{2}=x^{2}\) are linearly independent solutions on the entire real line of the equation $$ x^{2} y^{\prime \prime}-2 x y^{\prime}+2 y=0 $$ but that \(W\left(x, x^{2}\right)\) vanishes at \(x=0\). Why do these observations not contradict part (b) of Theorem 3 ?
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