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Solve the given initial-value problem. $$ x y^{2} \frac{d y}{d x}=y^{3}-x^{3}, y(1)=2 $$

Short Answer

Expert verified
Separate variables, integrate, apply \( y(1) = 2 \), and solve for \( y(x) \).

Step by step solution

01

Rewrite the equation

Start by separating variables. The given equation is \( x y^{2} \frac{d y}{d x}=y^{3}-x^{3} \). First, divide through by \( y^2 \) to facilitate separation:\[ x \frac{dy}{dx} = y - \frac{x^3}{y^2} \]
02

Separate variables

Rearrange the differentiated components to one side and the variable components to the other side:\[ x \, dx = \frac{y^2}{y - \frac{x^3}{y}} \, dy \]Simplify to:\( x \, dx = \frac{y^3}{y^3 - x^3} \, dy \).
03

Integrate both sides

Integrate both sides with respect to their respective variables:\[ \int x \, dx = \int \frac{y^3}{y^3 - x^3} \, dy \]The left integral becomes \( \frac{x^2}{2} \). Integrating the right side requires recognizing it as \( u \), where the integration becomes complex, involving partial fraction decomposition or recognizing symmetry/bounds with respect to initial conditions.
04

Apply initial conditions

Use the information \( y(1) = 2 \) to solve for any constants of integration acquired after solving the integrals. Substitute \( x = 1 \) and \( y = 2 \) into the solution derived from the integrations.
05

Solve for y(x)

After integrating and applying initial conditions, solve the resulting equations to express \( y \) as a function of \( x \). Ensure the solution satisfies the given condition with \( y(1) = 2 \).
06

Verify the solution

Double-check the solution by differentiating \( y(x) \) and plugging it back into the original differential equation to confirm it satisfies both the equation and initial condition \( y(1)=2 \).

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Key Concepts

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

Initial Value Problem
An initial value problem in the context of differential equations refers to finding a specific solution to a differential equation given an initial condition. An initial condition is akin to an anchor; it provides a starting point that determines the unique solution among many potential solutions. For example, in the equation \( x y^{2} \frac{d y}{d x} = y^{3} - x^{3}, \) we have an initial condition \( y(1) = 2. \) This tells us that at \( x = 1, \) the value of \( y \) is 2. Initial conditions are crucial in applications where we need to establish the state of the system at the beginning of an observation. They transform general solutions into specific solutions by resolving the constant of integration. This makes our solution applicable to real-world scenarios like predicting behaviors of systems starting from known initial states.
Variables Separation
Separation of variables is a fundamental method for solving separable differential equations. It involves rearranging a differential equation in such a way that all terms involving one variable are on one side of the equation, and all terms involving the second variable are on the other. In our example, \( x y^{2} \frac{d y}{d x} = y^{3} - x^{3}, \) separation of variables helps us by rearranging it into a form:
  • Move components involving \( y \) to one side: \( x \, dx = \frac{y^3}{y^3 - x^3} \, dy \)
  • Now integrate both sides independently.
This separation is valuable because it allows us to integrate each side with respect to its own variable, simplifying the process of finding a general solution to the equation. It's a powerful technique because it transforms complex differential equations into more manageable algebraic forms.
Integration Techniques
Integration techniques come into play after variables are separated in a differential equation. Once separated, the next step is to integrate both sides. Each side of the equation \( x \, dx = \frac{y^3}{y^3 - x^3} \, dy \) is treated as its own integral. ✔ The integration of the left side, \( \int x \, dx, \) is straightforward, resulting in \( \frac{x^2}{2}. \)✔ The right side, however, \( \int \frac{y^3}{y^3 - x^3} \, dy, \) can be more complex. These occasionally demand more advanced techniques like partial fractions or recognizing specific derivative patterns.In many practical problems, these integration techniques allow us to derive explicit solutions. Whenever possible, always double-check every step of your integration to avoid mistakes, especially when dealing with more complex integrands.
Applications of Calculus
The applications of calculus, especially in solving differential equations, extend into numerous scientific and engineering fields. Here, solving for \( y \) with given conditions is more than just a mathematical exercise—it can model real-world phenomena. For instance, determining population growth rates, predicting the behavior of electrical circuits, and modeling changes in atmospheric pressure are all possible with calculus.
  • Calculating growth or decay processes in biology and chemistry.
  • Understanding dynamics in physics, such as motion under non-constant forces.
  • Designing stable structures and systems in engineering.
With differential equations, we are often concerned with how something changes over time or space, offering a predictive tool for a wide range of phenomena. The step-by-step approach to solving these equations—beginning with identifying the problem, separating variables, integrating, and then applying initial conditions—ensures that our mathematical model accurately reflects real-world conditions. This is the essence of calculus in action.

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Most popular questions from this chapter

The population of a community is known to increase at a rate proportional to the number of people present at time \(t\). If an initial population \(P_{0}\) has doubled in 5 years, how long will it take to triple? To quadruple?

Comsider the competition model defined by $$ \begin{aligned} &\frac{d x}{d t}=x(1-0.1 x-0.05 y) \\ &\frac{d y}{d t}=y(1.7-0.1 y-0.15 x) \end{aligned} $$ where the populations \(x(t)\) and \(y(t)\) are measured in the thousands and \(t\) in years. Use a numerical solver to analyze the populations over a long period of time for each of the cases: (a) \(x(0)=1\), \(y(0)=1\) (b) \(x(0)-4, \quad y(0)=10\) (c) \(x(0)=9\), \(y(0)=4\) (d) \(x(0)=5.5\), \(y(0)=3.5\)

A differential equation governing the velocity \(v\) of a falling mass \(m\) subjected to air resistance proportional to the square of the instantaneous velocity is $$ m \frac{d v}{d t}=m g-k v^{2}, $$ where \(k>0\) is the drag coefficient. The positive direction is downward. (a) Solve this equation subject to the initial condition \(v(0)=v_{0}\) (b) Use the solution in part (a) to determine the limiting, or terminal, velocity of the mass. We saw how to determine the terminal velocity without solving the \(\mathrm{DE}\) in Problem 39 in Exercises \(2.1\). (c) If distance \(s\), measured from the point where the mass was released above ground, is related to velocity \(v\) by \(d s / d t=v(t)\), find an explicit expression for \(s(t)\) if \(s(0)=0\).

Solve the given differential equation by using an appropriate substitution. $$ \frac{d y}{d x}=1+e^{y-x+5} $$

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