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Magazine Chapter 7: Problem 1
Find the equation of the solution to \(\frac{d y}{d x}=x^{2} y\) through the point \((x, y)=(1,3)\). help (equations)
Short Answer
Expert verified
The equation of the solution is \(y = 3 e^{\frac{x^3 - 1}{3}}\).
Step by step solution
01
Identify the Type of Differential Equation
The given differential equation is \(\frac{d y}{d x}=x^{2} y\). This is a first-order linear ordinary differential equation (ODE) in the standard form.
02
Separate Variables
Rewrite the equation to separate the variables y and x: \(\frac{d y}{y} = x^2 ~ d x\).
03
Integrate Both Sides
Integrate both sides of the equation: \(\begin{aligned} \int \frac{1}{y} ~ d y & = \int x^2 ~ d x \end{aligned}\).
04
Solve the Integrals
Calculate the integrals: \(\begin{aligned} \ln|y| & = \frac{x^3}{3} + C \end{aligned}\), where C is the constant of integration.
05
Exponentiate Both Sides
To solve for y, exponentiate both sides to remove the natural logarithm: \(|y| = e^{\frac{x^3}{3} + C}\). Simplify this to \(|y| = e^{\frac{x^3}{3}} \times e^C\). Let \(e^C = k\), where k is a constant.
06
Include the Constant
Rewrite the solution: \(y = k e^{\frac{x^3}{3}}\).
07
Apply Initial Condition
Use the initial condition \((x, y) = (1, 3)\) to find the value of k. Substitute into the equation: \(3 = k e^{\frac{1^3}{3}}\). Simplify this to \(3 = k e^{\frac{1}{3}}\).
08
Solve for the Constant k
Solve for k: \(k = 3 e^{-\frac{1}{3}}\).
09
Write the Final Solution
Substitute the value of k back into the equation to get the final solution: \(y = 3 e^{-\frac{1}{3}} e^{\frac{x^3}{3}} = 3 e^{\frac{x^3 - 1}{3}} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
First-order differential equations
A first-order differential equation is an equation involving a function and its first derivative. In our case, the given differential equation is \ \(\frac{d y}{d x}=x^2 y\ \). This is known as a first-order ordinary differential equation (ODE) because it only contains the first derivative of the function with respect to the independent variable, which is x. These types of equations often represent real-world phenomena like population growth or the rate at which water drains from a tank. The goal is to find the function y in terms of x that satisfies this equation.
Separation of variables
Separation of variables is a straightforward method to solve first-order differential equations. The idea is to rearrange the equation so that each variable appears on a different side. For our differential equation \ \(\frac{d y}{d x}=x^2 y\ \), we can rewrite it to separate y and x. By dividing both sides by y and multiplying both sides by dx, we get: \ \(\frac{d y}{y} = x^2 ~ d x\ \). This allows us to integrate each side independently, which simplifies finding the general solution.
Integration of equations
Once the variables are separated, we integrate both sides of the equation to solve it. For our separated equation \ \(\frac{d y}{y} = x^2 ~ d x\ \), we integrate each side: \ \(\begin{aligned} \ \ \int \frac{1}{y} ~ d y & = \int x^2 ~ d x \ \ \end{aligned}\ \). The left side integrates to \ \(\begin{aligned} \ln|y| \end{aligned}\ \), while the right side integrates to \ \(\begin{aligned} \frac{x^3}{3} + C \end{aligned}\ \), where C is the constant of integration. Thus, we have the intermediary solution: \ \(\begin{aligned} \ln|y| &= \frac{x^3}{3} + C \end{aligned}\ \).
Initial value problem
An initial value problem involves solving a differential equation with an initial condition that specifies the value of the function at a particular point. In our exercise, the initial condition is given by the point \ \((x, y) = (1,3)\ \). To find the final solution, we use this initial condition to determine the constant of integration (k). Substitute the initial conditions into the general solution \ \( y = k e^{\frac{x^3}{3}} \ \): \ \(3 = k e^{\frac{1}{3}} \ \). Solving for k, we get \ \( k = 3 e^{-\frac{1}{3}} \ \). Hence, the final solution is: \ \( y = 3 e^{\frac{x^3-1}{3}} \ \), which satisfies both the differential equation and the initial condition.
