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Find the general solution of each differential equation or state that the differential equation is not separable. If the exercise says "and check," verify that your answer is a solution. $$ y^{\prime}=x^{m} y \quad \text { (for } \left.m>0\right) $$

Short Answer

Expert verified
The general solution is \( y = Ae^{\frac{x^{m+1}}{m+1}} \).

Step by step solution

01

Identify the Type of Differential Equation

The given differential equation is \( y' = x^m y \). This is a first-order differential equation, and it's separable because we can express it in the form \( \frac{dy}{dx} = x^m y \).
02

Separate Variables

Rewrite the equation to separate the variables, \( y \) on one side and \( x \) on the other. We have:\[ \frac{dy}{y} = x^m dx \].
03

Integrate Both Sides

Integrate both sides of the equation separately. Integrating the left side with respect to \( y \) gives \( \ln|y| \) and the right side with respect to \( x \) gives:\[ \ln|y| = \frac{x^{m+1}}{m+1} + C \] where \( C \) is the constant of integration.
04

Solve for \( y \)

To solve for \( y \), exponentiate both sides to remove the natural logarithm:\[ |y| = e^{\left(\frac{x^{m+1}}{m+1} + C\right)} \], which simplifies to \( y = Ae^{\frac{x^{m+1}}{m+1}} \), where \( A = e^C \) is an arbitrary constant.
05

General Solution

The general solution to the differential equation is: \[ y = Ae^{\frac{x^{m+1}}{m+1}} \], where \( A \) is any constant.

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Key Concepts

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

Separable Differential Equations
Separable differential equations are a special type of first-order differential equation. These can be expressed in the form \( \frac{dy}{dx} = g(x)h(y) \), where we can separate the variables allowing \( x \) terms and \( y \) terms to be on different sides of the equation. This form is beneficial because it allows us to simplify the equation significantly, making integration possible.
By separating variables, we can rewrite the equation to solve for the function \( y \) in terms of \( x \). This process involves moving all terms involving \( y \) (including \( dy \)) to one side and all terms involving \( x \) (including \( dx \)) to the other side.
This method is crucial for many problems because it paves the way to solving the equation through integration, making the solution procedure more straightforward.
Integration
Integration is a fundamental concept used to solve differential equations. Once variables are separated, integration helps find the function or functions satisfying the differential equation. This process comes after rearranging terms in a separable differential equation.
The idea of integration involves finding the antiderivative or integral of the separated terms. For instance, if you have an expression like \( \int \frac{dy}{y} = \int x^m dx \), you integrate both sides independently.
This results in integrating \( \frac{1}{y} \) with respect to \( y \) yielding \( \ln|y| \), and \( x^m \) with respect to \( x \) resulting in \( \frac{x^{m+1}}{m+1} \). Therefore, integration plays a key role in finding general solutions to separable differential equations.
General Solution
A general solution of a differential equation represents a set of all possible solutions that satisfy the equation. It usually involves an arbitrary constant, which can be adjusted to find particular solutions.
For our example, after integrating and re-arranging terms, the general solution for the differential equation \( y' = x^m y \) is \( y = Ae^{\frac{x^{m+1}}{m+1}} \). Here, \( A \) is an arbitrary constant determined by the initial conditions or additional constraints if provided.
The presence of \( A \) means the solution covers a family of curves depending on the value of \( A \), representing different possible behaviors of the solution.
First-Order Differential Equation
First-order differential equations are equations which contain only the first derivative \( y' \) with respect to the independent variable, usually \( x \). First-order equations are simpler to solve than higher-order ones, as they involve finding one integral.
These types of equations can take various forms, including separable, linear, and exact, among others. The examaple \( y' = x^m y \) is a first-order equation as it includes only the derivative with respect to \( x \) and not higher order derivatives.
Understanding the form of the first-order differential equation helps in determining the appropriate method for finding solutions, such as the separation of variables technique used in this example.

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Most popular questions from this chapter

When you swallow a pill, the medication passes through your stomach lining into your bloodstream, where some is absorbed by the cells of your body and the rest continues to circulate for future absorption. The amount \(y(t)\) of medication remaining in the bloodstream after \(t\) hours can be modeled by the differential equation $$ \frac{d y}{d t}=a b e^{-b t}-c y $$ for constants \(a, b,\) and \(c\) (respectively the dosage of the pill, the dissolution constant of the pill, and the absorption constant of the medication). For the given values of the constants: a. Substitute the constants into the stated differential equation. b. Solve the differential equation (with the initial condition of having no medicine in the bloodstream at time \(t=0)\) to find a formula for the amount of medicine in the bloodstream at any time \(t\) (hours). c. Use your solution to find the amount of medicine in the bloodstream at time \(t=2\) hours. d. Graph your solution on a graphing calculator and find when the amount of medication in the bloodstream is maximized. \(a=30 \mathrm{mg}, \quad b=5, \quad c=0.3\)

Solve each differential equation with the given initial condition. $$ \begin{array}{l} y^{\prime}+2 x y=4 x \\ y(0)=0 \end{array} $$

Solve each by the appropriate technique. a. \(y y^{\prime}=x\) b. \(y^{\prime}+y=e^{-x}\)

Solve each differential equation and initial condition and verify that your answer satisfies both the differential equation and the initial condition. $$\left\\{\begin{array}{l} y^{\prime}=2 \sqrt{y} \\ y(1)=4 \end{array}\right.$$

Suppose that you meet 30 new people each year, but each year you forget \(20 \%\) of all of the people that you know. If \(y(t)\) is the total number of people who you remember after \(t\) years, then \(y\) satisfies the differential equation \(y^{\prime}=30-0.2 y .\) (Do you see why?) Solve this differential equation subject to the condition \(y(0)=0\) (you knew no one at birth).

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