/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 2 Check that \(y=t^{4}\) is a solu... [FREE SOLUTION] | 91Ó°ÊÓ

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

Short Answer

Expert verified
Yes, \(y = t^4\) is a solution to the differential equation.

Step by step solution

01

Differentiate the given function

Start by finding the derivative of the given function with respect to \(t\). For \(y = t^4\), the derivative \( \frac{d y}{d t} \) can be found using the power rule: \( \frac{d y}{d t} = 4t^{3} \).
02

Substitute derivatives into the differential equation

Substitute \(y = t^4\) and \(\frac{d y}{d t} = 4t^3\) into the given differential equation \(t \frac{d y}{d t} = 4 y\). This gives: \( t \cdot 4t^3 = 4 \cdot t^4 \).
03

Simplify the equation

Simplify the expression from Step 2. Calculate \(t \cdot 4t^3 = 4t^4\) and recognize that \(4 \cdot t^4 = 4t^4\).
04

Confirm that the equation holds

After simplification, observe that both sides of the equation are equal: \(4t^4 = 4t^4\). This confirms that \(y = t^4\) satisfies the differential equation \(t \frac{d y}{d t} = 4 y\).

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Key Concepts

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

Solutions Verification
To verify if a function is indeed a solution to a differential equation, start by ensuring that all operations and substitutions are executed correctly. This involves confirming that the function, once placed into the differential equation, makes both sides of the equation equal each other. You begin by determining if the expression satisfies all parts of the equation when inserted as indicated.

A typical process for solutions verification includes:
  • Substitution: Plug the given function and its derivative into the equation.
  • Check Consistency: Simplify both sides of the differential equation to confirm they are identical.
In our example, we inserted the derived function, integrated it within the differential expression, and after simplification, both sides matched, confirming it as a valid solution.
Power Rule Differentiation
Differentiation is a fundamental aspect in calculus, used to find the rate of change of a function with respect to one of its variables. The power rule is a basic differentiation rule applied to functions of the form, where a variable is raised to a constant power.

The power rule states that if you have a term, say, \( t^n \), the derivative \( \frac{d}{dt}(t^n) \) is calculated as \( n \, t^{n-1} \).
For example, let's differentiate \( y = t^4 \):
  • Identify the exponent: Here, \( n = 4 \).
  • Apply the power rule: The derivative becomes \( 4t^3 \) because \( 4 \cdot t^{4-1} = 4t^3 \).
The application of the power rule in this problem demonstrates how simple mathematical rules can be used effectively to arrive at solutions, prepared for subsequent steps like substitution.
Substitution in Equations
Substitution is a key step in solving differential equations. It involves replacing variables and functions with their actual values or expressions, helping simplify the equation for easier analysis.

In the process of solving the differential equation, we substituted both the original function and its derivative:
  • Original function substitution: Replace \( y \) with \( t^4 \).
  • Derivative substitution: Replace \( \frac{dy}{dt} \) with \( 4t^3 \).
This method helps the equation reach a form where further simplification can show that both sides of the equation are balanced. By putting the expressions into the equation \( t \frac{d y}{d t} = 4y \), the substition reflects initial condition making the later simplification clearer: \( t \cdot 4t^3 = 4 \cdot t^4 \), showing equality, hence proving the function is a solution.

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Most popular questions from this chapter

Find solutions to the differential equations in subject to the given initial condition. $$\frac{d Q}{d t}=\frac{Q}{5}, \quad Q=50 \text { when } t=0$$

(a) Consider the slope field for \(d y / d x=x y .\) What is the slope of the line segment at the point (2,1)\(?\) At (0,2)\(?\) At (-1,1)\(?\) At (2,-2)\(?\) (b) Sketch part of the slope field by drawing line segments with the slopes calculated in part (a).

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

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, \(t\)

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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