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Chapter 6: Problem 1

Find the general solution of the differential equation. $$ \frac{d y}{d x}=\frac{x}{y} $$

Short Answer

Expert verified
The general solution of the differential equation \(\frac{d y}{d x}=\frac{x}{y}\) is \(y = \pm \sqrt{2(x+C)}\).

Step by step solution

01

Recognize type of differential equation

Firstly, identify that the given differential equation \(\frac{d y}{d x} = \frac{x}{y}\) is a separable equation. This is because it can be written in the separable form \(g(y) dy = f(x) dx\).
02

Separate variables

Rearrange the equation so that all terms involving y are on one side and all terms involving x are on the other side. \(\frac{y}{x} dy = dx\).
03

Integrate both sides

Now, having separated the variables, integrate both sides with respect to their own variables. The integral of the left side with respect to y will be \(\int y dy\) and the integral of the right side with respect to x will be \(\int dx\). This gives \(0.5y^2 = x + C\), where C is the constant of integration.
04

Express y in terms of x

Finally, solving \(0.5y^2 = x + C\) for y, the general solution of the given differential equation is obtained as \(y = \pm \sqrt{2(x+C)} \).

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Key Concepts

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

Integration
Integration is a fundamental concept in calculus used to determine the accumulation of quantities. It's essentially the reverse of differentiation. In the context of solving separable differential equations, integration is used to solve for the unknown function given its derivative.
In this exercise, once we separate the variables to get \(\frac{y}{x} dy = dx\), we integrate both sides to help solve the differential equation. This means we find the anti-derivative for each side:
  • For \(y dy\), the integral is \(\frac{1}{2}y^2\)
  • For \(dx\), the integral is simply \(x\) since it represents a constant derivative
After integration, we end up with the equation \(\frac{1}{2}y^2 = x + C\). This integration step is vital as it transforms the differential equation into a form that is often simpler and solvable.
Constant of Integration
When we integrate a function, we should always consider the constant of integration \(C\). This constant arises because when differentiating a constant, it becomes zero, leading to the possibility of infinitely many antiderivatives. Therefore, to account for all forms that the integral could take, \(C\) is added.
In our step-by-step solution, we see \(C\) appear on the right side of the equation after integrating: \(\frac{1}{2}y^2 = x + C\). \
  • The constant of integration \(C\) represents the family of solutions for a differential equation.
  • Each different value of \(C\) results in a different particular solution to the differential equation.
Understand that \(C\) is crucial for capturing all possible solutions of the integral, especially when considering initial conditions or specific contexts.
General Solution
The general solution of a differential equation includes not just one, but all possible solutions. This is because it incorporates the constant of integration, \(C\), representing the different values that can satisfy the equation.
In our exercise, the final step gives us the general solution \(y = \pm \sqrt{2(x+C)}\). This expression indicates that:
  • There are two branches (positive and negative) for \(y\), due to the square root function.
  • \(C\) is any real number, showing that the equation solution isn't unique without additional conditions.
The general solution is important as it provides a comprehensive answer to the differential equation, encompassing every scenario and initial condition that might apply in real-world situations.

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Most popular questions from this chapter

Solve the Bernoulli differential equation. $$ y^{\prime}+\left(\frac{1}{x}\right) y=x y^{2} $$

Determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false. The differential equation modeling exponential growth is \(d y / d x=k y\), where \(k\) is a constant.

Solve the first-order differential equation by any appropriate method. $$ \left(2 y-e^{x}\right) d x+x d y=0 $$

A calf that weighs 60 pounds at birth gains weight at the rate \(\frac{d w}{d t}=k(1200-w)\) where \(w\) is weight in pounds and \(t\) is time in years. Solve the differential equation. (a) Use a computer algebra system to solve the differential equation for \(k=0.8,0.9\), and 1 . Graph the three solutions. (b) If the animal is sold when its weight reaches 800 pounds, find the time of sale for each of the models in part (a). (c) What is the maximum weight of the animal for each of the models?

Find the time necessary for \( 1000\) to double if it is invested at a rate of \(r\) compounded (a) annually, (b) monthly, (c) daily, and (d) continuously. $$ r=5.5 \% $$
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