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Chapter 9: Problem 21

Find the values of \(k\) for which \(y=x^{2}+k\) is a solution to the differential equation \(2 y-x y^{\prime}=10\)

Short Answer

Expert verified
The value of \(k\) is 5.

Step by step solution

01

Identify the given functions

The function we need to check as a solution is \(y = x^2 + k\). This will be our function that we will differentiate and test in the differential equation.
02

Differentiate the function y

Compute the derivative of the function \(y = x^2 + k\). The derivative \(y'\) with respect to \(x\) is \(y' = 2x\).
03

Substitute y and y' into the differential equation

Substitute \(y = x^2 + k\) and \(y' = 2x\) into the differential equation \(2y - xy' = 10\). After substitution, the equation becomes \(2(x^2 + k) - x(2x) = 10\).
04

Simplify the substituted equation

Simplify \(2(x^2 + k) - x(2x) = 10\) to solve for \(k\). This leads to \(2x^2 + 2k - 2x^2 = 10\), simplifying to \(2k = 10\).
05

Solve for k

Divide both sides of the equation \(2k = 10\) by 2 to solve for \(k\), resulting in \(k = 5\). Thus, \(k\) must be 5 for the function to be a solution.

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Key Concepts

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

Derivative
Differentiating a function involves calculating its rate of change relative to its variable. This process reveals how the function behaves as its input values fluctuate. For the function \(y = x^2 + k\), the derivative is found by examining its components. The term \(x^2\) derives to \(2x\), while the constant \(k\) vanishes since its rate of change is zero. Therefore, the derivative \(y' = 2x\) provides essential information for solving differential equations. Differentiation highlights how sensitive the function is to changes in its input variable \(x\). By realizing that \(y' = 2x\), we prepare to incorporate it later in our equation-solving process.
  • Differentiate terms individually
  • Constants derive to zero
  • Derivatives provide rate of change
Substitution
Substitution in mathematics involves replacing variables with corresponding values or expressions. In our differential equation case, we substitute derived expressions back into the original equation to inspect if they satisfy it. Our function \(y = x^2 + k\) and its derivative \(y' = 2x\) are positioned within the differential equation \(2y - xy' = 10\). This substitution transforms the equation into a format that reflects the dependency of \(x\) and \(k\). Performing this step ensures we correctly align our substitutions with our equation, ultimately aiding our solution process.
  • Use substitution for testing solutions
  • Ensure correct placement in equations
  • Essential for deriving consistent results
Algebraic Simplification
Algebraic simplification is crucial for solving complex equations. It involves reducing expressions to their simplest forms to understand or solve them better. Starting with the equation \(2(x^2 + k) - x(2x) = 10\), we simplify by carrying out basic arithmetic operations. Upon expansion and rearrangement, components such as \(2x^2\) cancel each other. This reduction process shows that \(2k = 10\), a much simpler expression. Simplifying equations often leads to insights that are less visible in their initial forms.
  • Simplify to uncover simpler forms
  • Identify and cancel similar terms
  • Essential for clearly understanding results
Calculus Problem Solving
Solving calculus problems often involves a systematic approach involving several key steps. Beginning with identifying and differentiating the function, calculating the derivative, performing substitutions into a given differential equation, followed by algebraic simplifications to arrive at a solution. In our situation with the function \(y = x^2 + k\), we followed these steps to find \(k\) by first deriving, then substituting and simplifying the problem's equation. This structured pathway from calculus helps thoroughly examine variable behaviors under different conditions, ensuring precision and clarity in the solutions.
  • Structured problem-solving pathway
  • Combines differentiation, substitution, and simplification
  • Optimizes understanding of variable interactions

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Most popular questions from this chapter

Toxins in pesticides can get into the food chain and accumulate in the body. A person consumes 10 micrograms a day of a toxin, ingested throughout the day. The toxin leaves the body at a continuous rate of \(3 \%\) every day. Write a differential equation for the amount of toxin, \(A\) in micrograms, in the person's body as a function of the number of days, \(t\)

Some people write the solution of the initial value problem $$\frac{d y}{d t}=k(y-A) \quad y=y_{0} \text { at } t=0$$ in the form $$\frac{y-A}{y_{0}-A}=e^{k t}$$ Show that this formula gives the correct solution for \(y\) assuming \(y_{0} \neq A\).

Find particular solutions. $$\frac{d y}{d t}=0.5(y-200), \quad y=50\( when \)t=0$$

Check that \(y=t^{4}\) is a solution to the differential equation \(t \frac{d y}{d t}=4 y\)

The following system of differential equations represents the interaction between two populations, \(x\) and \(y\) $$\begin{aligned} &\frac{d x}{d t}=-3 x+2 x y\\\ &\frac{d y}{d t}=-y+5 x y \end{aligned}$$ (a) Describe how the species interact. How would each species do in the absence of the other? Are they helpful or harmful to each other? (b) If \(x=2\) and \(y=1,\) does \(x\) increase or decrease? Does \(y\) increase or decrease? Justify your answers. (c) Write a differential equation involving \(d y / d x\) (d) Use a computer or calculator to draw the slope field for the differential equation in part (c). (e) Draw the trajectory starting at point \(x=2, y=1\) on your slope field, and describe how the populations change as time increases.
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