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Chapter 6: Problem 39

Solve each differential equation and initial condition and verify that your answer satisfies both the differential equation and the initial condition. $$ \left\\{\begin{array}{l} y^{\prime}=x y \\ y(0)=-1 \end{array}\right. $$

Short Answer

Expert verified
The solution is \( y = -e^{\frac{x^2}{2}} \), which satisfies both the differential equation and the initial condition.

Step by step solution

01

Understand the Differential Equation

We are given a first-order linear differential equation \( y' = xy \). We need to solve this equation subject to the initial condition \( y(0) = -1 \).
02

Use Separation of Variables

To solve \( y' = xy \), separate the variables: \( \frac{dy}{y} = x \, dx \). The goal is to separate the variables into two integrals.
03

Integrate Both Sides

Integrate both sides of \( \frac{dy}{y} = x \, dx \). The integral of \( \frac{1}{y} \, dy \) is \( \ln|y| \) and the integral of \( x \, dx \) is \( \frac{x^2}{2} \). This gives: \( \ln|y| = \frac{x^2}{2} + C \), where \( C \) is the constant of integration.
04

Solve for \( y \)

Exponentiate both sides to solve for \( y \): \( |y| = e^{\frac{x^2}{2} + C} = e^{C}e^{\frac{x^2}{2}} \). Let \( A = e^{C} \), so \( |y| = A e^{\frac{x^2}{2}} \). Therefore, \( y = \pm A e^{\frac{x^2}{2}} \).
05

Apply the Initial Condition

Use the initial condition \( y(0) = -1 \) to determine \( A \). Substitute \( x = 0 \) and \( y = -1 \) into \( y = \pm A e^{\frac{x^2}{2}} \), giving \(-1 = \pm A e^{0} = \pm A \). Thus, \( A = -1 \).
06

Write the Particular Solution

The particular solution satisfying the initial condition is \( y = -e^{\frac{x^2}{2}} \).
07

Verify the Differential Equation

Differentiate \( y = -e^{\frac{x^2}{2}} \) with respect to \( x \) to find \( y' \). We have \( y' = -e^{\frac{x^2}{2}} \cdot x = x(-e^{\frac{x^2}{2}}) = xy \). Since \( y' = xy \) satisfies the given differential equation, the solution is verified.

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Key Concepts

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

Separation of Variables
Separation of variables is a method used to solve differential equations, particularly those that can be written in the form \( y' = f(x)g(y) \). The key idea is to "separate" the variables \( x \) and \( y \) onto opposite sides of the equation. The process involves:
  • Rewriting the equation so that all terms involving \( y \) are on one side (with \( dy \)), and all terms involving \( x \) are on the other side (with \( dx \)). For example, in the given equation \( y' = xy \), this becomes \( \frac{dy}{y} = x \, dx \).
  • Integrating both sides with respect to their respective variables. This involves calculus operations such as computing \( \int \frac{1}{y} \, dy \) for the left side, and \( \int x \, dx \) for the right side.
Once integrated, we solve for the function \( y \). This method is particularly useful because it transforms a differential equation into two simpler integral equations.
Initial Condition
An initial condition in solving differential equations is essentially the starting point at which a solution is known or specified. It allows us to find the specific, rather than a general solution. When solving an equation, like \( y' = xy \), we might end up with a family of solutions parameterized by a constant \( C \).
  • At \( x = 0 \), we had the condition that \( y = -1 \). This condition is critical in determining the exact value of the constant \( C \) (or in this case \( A \) since \( C \) is transformed into \( A = e^C \)).
  • By substituting \( x = 0 \) and \( y = -1 \), we could solve \( -1 = \pm A \) to find that \( A = -1 \).
Thus, initial conditions are essential to nail down a unique solution from all possible solutions of a differential equation.
Integrating Factor
Integrating factor is a technique used in solving linear first-order differential equations of the form \( y' + p(x)y = q(x) \). Although not used in this particular solution, it's helpful to understand that:
  • The integrating factor modifies an equation into an easily integrable form.
  • It is usually given as \( \mu(x) = e^{\int p(x) \, dx} \).
The method works by:
  • Multiplying the entire differential equation by this integrating factor \( \mu(x) \), transforming it into a form that allows the left-hand side to become the derivative of a product.
  • This simplification helps directly integrate both sides with ease, often leading to a solution for \( y \).
Though not applied here directly, integrating factors are a powerful tool when separation of variables is not straightforward.

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Most popular questions from this chapter

$$ \text { Use integration by parts to find each integral. } $$ $$ \int \sqrt[3]{x} \ln x d x $$

Evaluate each definite integral using integration by parts. (Leave answers in exact form.) $$ \int_{0}^{4} z(z-4)^{6} d z $$

Let \(y(t)\) be the size of a colony of bacteria after \(t\) hours. a. Write a differential equation that says that the rate of growth of the colony is equal to eight times the three-fourths power of its present size. b. Write an initial condition that says that at time zero the colony is of size 10,000 . c. Solve the differential equation and initial condition. d. Use your solution to find the size of the colony at time \(t=6\) hours.

Solve each differential equation and initial condition and verify that your answer satisfies both the differential equation and the initial condition. $$ \left\\{\begin{array}{l} y^{\prime}=x y-5 x \\ y(0)=4 \end{array}\right. $$

We omit the constant of integration when we integrate \(d v\) to get \(v\). Including the constant \(C\) in this step simply replaces \(v\) by \(v+C\), giving the formula $$ \int u d v=u(v+C)-\int(v+C) d u $$ Multiplying out the parentheses and expanding the last integral into two gives $$ \int u d v=u v+C u-\int v d u-C \int d u $$ Show that the second and fourth terms on the right cancel, giving the "old" integration by parts formula \(\int u d v=u v-\int v d u\). This shows that including the constant in the \(d v\) to \(v\) step gives the same formula. One constant of integration at the end is enough. Repeated Integration by Parts Sometimes an integral requires two or more integrations by parts. As an example, we apply integration by parts to the integral \(\int x^{2} e^{x} d x\). $$ \begin{array}{c} \int \underbrace{x^{2} e^{x} d x}_{u d v}=\underbrace{x^{2} e^{x}}_{u v}-\int \underbrace{e^{x} 2 x d x}_{w \atop d u}=x^{2} e^{x}-2 \int x e^{x} d x \\ {\left[\begin{array}{cc} u=x^{2} & d v & =e^{x} d x \\ d u=2 x d x & v=\int e^{x} d x=e^{x} \end{array}\right]} \end{array} $$ The new integral \(\int x e^{x} d x\) is solved by a second integration by parts. Continuing with the previous solution, we choose new \(u\) and \(d u\) :
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