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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. $$ y^{\prime}=12 x^{3} y \quad \text { and check } $$

Short Answer

Expert verified
The general solution is \( y = Ce^{3x^4} \).

Step by step solution

01

Identify the Type of Differential Equation

The given differential equation is \( y' = 12x^3y \). This is a first-order differential equation of the form \( y' = g(x)h(y) \), which suggests it may be separable. A separable equation can be expressed as \( \frac{dy}{dx} = g(x)h(y) \).
02

Rewrite the Equation in Separable Form

Rewrite the equation \( y' = 12x^3y \) as \( \frac{dy}{dx} = 12x^3y \). This can be rearranged into \( \frac{dy}{y} = 12x^3 \, dx \). This shows that the differential equation is separable.
03

Integrate Both Sides

Integrate both sides of \( \frac{dy}{y} = 12x^3 \, dx \). The integration gives \( \int \frac{dy}{y} = \int 12x^3 \, dx \). The result is \( \ln |y| = 3x^4 + C \), where \( C \) is the constant of integration.
04

Solve for \( y \)

Exponentiate both sides of \( \ln |y| = 3x^4 + C \) to solve for \( y \). This gives \( |y| = e^{3x^4 + C} \). Simplify it to \( y = Ce^{3x^4} \), where \( C \) is a constant, since \( e^C \) is simply another constant.
05

Check the Solution

Differentiate \( y = Ce^{3x^4} \) to verify that it satisfies the original differential equation. The derivative is \( y' = C \cdot 12x^3 e^{3x^4} \). Since \( y = Ce^{3x^4} \), we have \( y' = 12x^3y \), which matches the original differential equation. Therefore, the solution is verified.

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Key Concepts

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

First-Order Differential Equations
First-order differential equations are equations that have only the first derivative of a function. These do not include higher derivatives like the second or third derivative. A typical form of a first-order differential equation is \( y' = f(x, y) \). Here, \( y' \) represents the first derivative of the unknown function \( y \) with respect to \( x \).

These equations are foundational in differential equations because they simplify the relationships between variables. They often model systems where the rate of change of the dependent variable is expressed in terms of the independent variable and the dependent variable itself.

To solve these equations, we frequently seek methods that isolate derivatives and simplify integration. You'll often find them categorized based on their terms' configurations, such as linear or non-linear, and further on whether they are homogenous or non-homogenous based on their relation to the zero function.
Separable Differential Equations
Separable differential equations are a special type of first-order differential equations where all terms containing the dependent variable can be separated from those containing the independent variable. This separation allows the equation to be easily solved by integration.

A differential equation of this form looks like \( \frac{dy}{dx} = g(x)h(y) \). Through algebraic manipulation, we can rewrite it as \( \frac{dy}{h(y)} = g(x)dx \). This separation sets the stage for integration on both sides to find a general solution.

One noteworthy aspect of separable equations is their application in real-world scenarios. They are used to model growth processes, chemical reactions, and other systems where current values define the rate of change. Recognizing separable equations quickly and transitioning them into an integrated format is key to solving many practical problems.
General Solution of Differential Equations
The general solution of a differential equation is the complete set of all possible solutions. For first-order differential equations, the general solution typically includes arbitrary constants that represent initial conditions or particular solutions.

For the problem \( y^{\prime}=12 x^{3} y \), rewriting it as \( \frac{dy}{y} = 12x^3 \, dx \) led to integrating both sides to attain \( \ln |y| = 3x^4 + C \). Solving for \( y \) gave us \( y = Ce^{3x^4} \), where \( C \) is the arbitrary constant representing any of the infinitely many solutions to this equation.

Understanding the general solution's significance is crucial because it includes all potential situations the equation could describe. Particular solutions are derived by applying initial conditions or boundary conditions to these general solutions, thus tailoring it to specific scenarios or constraints.

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Most popular questions from this chapter

Solve each differential equation with the given initial condition. $$ \begin{array}{l} x y^{\prime}=2 y+x^{2} \\ y(1)=3 \end{array} $$

Solve each differential equation with the given initial condition. $$ \begin{array}{l} y^{\prime}-3 x^{2} y=6 x^{2} \\ y(0)=1 \end{array} $$

A 500-gallon tank is filled with water containing 0.2 ounce of impurities per gallon. Each hour, 200 gallons of water (containing 0.01 ounce of impurities per gallon) is added and mixed into the tank, while an equal volume of water is removed. a. Write a differential equation and initial condition that describe the total amount \(y(t)\) of impurities in the tank after \(t\) hours. b. Solve this differential equation and initial condition. c. Use your solution to find when the impurities will reach 0.05 ounce per gallon, at which time the water may be used for drinking. d. Use your solution to find the "long-run" amount of impurities in the tank.

An annuity is a fund into which one makes equal payments at regular intervals. If the fund earns interest at rate \(r\) compounded continuously, and deposits are made continuously at the rate of \(d\) dollars per year (a continuous annuity), then the value \(y(t)\) of the fund after \(t\) years satisfies the differential equation \(y^{\prime}=d+r y .\) (Do you see why?) Solve the differential equation in the preceding instructions for the continuous annuity \(y(t),\) where \(d\) and \(r\) are unknown constants, subject to the initial condition \(y(0)=0\) (zero initial value).

The following exercises require the use of a slope field program. For each differential equation: a. Use a graphing calculator slope field program to graph the slope field for the differential equation on the window [-5,5] by [-5,5]. b. Sketch the slope field on a piece of paper and draw a solution curve that follows the slopes and that passes through the given point. $$ \begin{array}{l} \frac{d y}{d x}=\frac{x}{y^{2}+1} \\ \text { point: }(0,-1) \end{array} $$

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