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Chapter 3: Problem 72

Solve the differential equation subject to the indicated condition. \(\frac{d y}{d t}=t^{2} y^{4} ; y=1\) at \(t=1\)

Short Answer

Expert verified
The solution is \( y = (-t^3)^{-1/3} \).

Step by step solution

01

Separate Variables

To solve the differential equation \( \frac{dy}{dt} = t^2 y^4 \), start by separating the variables. Get all terms involving \(y\) on one side and all terms involving \(t\) on the other side. This gives you \( y^{-4} dy = t^2 dt \).
02

Integrate Both Sides

Integrate both sides of the equation. On the left, integrate \( \int y^{-4} dy \), and on the right, integrate \( \int t^2 dt \). The integral of \( y^{-4} \) is \( \frac{y^{-3}}{-3} + C_1 \), and the integral of \( t^2 \) is \( \frac{t^3}{3} + C_2 \). This yields:\[ -\frac{1}{3}y^{-3} = \frac{t^3}{3} + C_2. \]
03

Simplify and Solve for y

Multiply both sides by \(-3\) to simplify:\[ y^{-3} = -t^3 + C. \]Solve for \( y \) by taking the reciprocal to get:\[ y = \left(-t^3 + C\right)^{-1/3}. \]
04

Apply Initial Condition

Use the initial condition \( y = 1 \) when \( t = 1 \) to find the constant \( C \). Substituting, we have:\[ 1 = (-1^3 + C)^{-1/3}, \]which simplifies to \( 1 = (1 + C)^{-1/3} \). Solving for \( C \), \((1 + C) = 1\), so \( C = 0 \).
05

Write the Final Solution

Substitute \( C = 0 \) back into the equation for \( y \):\[ y = (-t^3 + 0)^{-1/3} = (-t^3)^{-1/3}. \]The solution is thus \( y = (-t^3)^{-1/3}. \)

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Key Concepts

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

Separation of Variables
In differential equations, separation of variables is a technique used to solve ordinary differential equations. The method involves rearranging the equation so that each variable appears on one side of the equation. For the equation \( \frac{dy}{dt} = t^2 y^4 \), this means moving all the terms involving \( y \) to one side and all the terms involving \( t \) to the other side.
  • The equation becomes \( y^{-4} dy = t^2 dt \), where each side of the equation now contains only one variable.
  • This separation makes it possible to integrate each side independently, which is a fundamental step in solving the equation.
The goal of this technique is to simplify the process of finding an antiderivative, which is a crucial part of solving differential equations.
Initial Condition
The initial condition is a vital component of solving differential equations as it allows us to determine the particular solution from a general solution. In the exercise, the initial condition is given as \( y = 1 \) at \( t = 1 \).
  • Using the initial condition helps in finding the constant of integration, which appears after integrating both sides of the separated equation.
  • In this case, substituting \( y = 1 \) and \( t = 1 \) into the equation \( y = (-t^3 + C)^{-1/3} \) allows us to solve for \( C \).
This process helps bridge the general solution to the specific scenario dictated by the problem, ensuring that the solution fits the real-world or theoretical constraints given in the problem.
Integration
Integration is a fundamental concept in calculus, especially when solving differential equations. Once variables have been separated, the next task is to integrate both sides of the equation. For this problem:
  • On the left side, we integrate \( \int y^{-4} dy \), which results in \( \frac{y^{-3}}{-3} + C_1 \).
  • On the right side, we integrate \( \int t^2 dt \), which gives \( \frac{t^3}{3} + C_2 \).
The integration process provides the antiderivatives or the general functions that satisfy the differential equation in question. This step is crucial for transforming the equation into a form that can be further solved, especially with the help of initial conditions.
Calculus
Calculus encompasses differentiation and integration, both of which are instrumental in solving differential equations. In this exercise, calculus principles guide us from separating variables through integration to applying the initial condition.
  • Differential equations like \( \frac{dy}{dt} = t^2 y^4 \) inherently involve derivatives, which are a core idea in calculus. Solving them requires a deep understanding of integration and differentiation.
  • Integration, a key process within calculus, allows us to find functions whose derivatives match the separated terms \( y^{-4} \) and \( t^2 \).
  • Calculus also teaches us to apply initial conditions correctly to find specific solutions from general solutions.
Overall, calculus provides the toolkit necessary to work through differential equations and find meaningful solutions in various contexts.

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Most popular questions from this chapter

Approximate the values of \(x\) that give maximum and minimum values of the function on the indicated intervals. $$ f(x)=\frac{\sin x}{x} ;[\pi, 3 \pi] $$

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In installment buying, one would like to figure out the real interest rate (effective rate), but unfortunately this involves solving a complicated equation. If one buys an item worth \(\$ P\) today and agrees to pay for it with payments of \(\$ R\) at the end of each month for \(k\) months, then $$ P=\frac{R}{i}\left[1-\frac{1}{(1+i)^{k}}\right] $$ where \(i\) is the interest rate per month. Tom bought a used car for \(\$ 2000\) and agreed to pay for it with \(\$ 100\) payments at the end of each of the next 24 months. (a) Show that \(i\) satisfies the equation $$ 20 i(1+i)^{24}-(1+i)^{24}+1=0 $$ (b) Show that Newton's Method for this equation reduces to $$ i_{n+1}=i_{n}-\left[\frac{20 i_{n}^{2}+19 i_{n}-1+\left(1+i_{n}\right)^{-23}}{500 i_{n}-4}\right] $$ (c) Find \(i\) accurate to five decimal places starting with \(i=0.012,\) and then give the annual rate \(r\) as a percent \((r=1200 i)\)

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\(f^{\prime \prime}(x)\) is given. Find \(f(x)\) by antidifferentiating twice. Note that in this case your answer should involve two arbitrary constants, one from each antidifferentiation. For example, if \(f^{\prime \prime}(x)=x,\) then \(f^{\prime}(x)=x^{2} / 2+C_{1}\) and \(f(x)=\) \(x^{3} / 6+C_{1} x+C_{2} .\) The constants \(C_{1}\) and \(C_{2}\) cannot be combined because \(C_{1} x\) is not a constant. $$ f^{\prime \prime}(x)=\frac{x^{4}+1}{x^{3}} $$
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