Skip to main content
Ch. 9 - Differential Equations
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
All textbooksBriggs 3rd EditionCh. 9 - Differential EquationsProblem 9.R.14
Chapter 9, Problem 9.R.14

11–18. Solving initial value problems Use the method of your choice to find the solution of the following initial value problems.
y′(x) = x/y, y(2) = 4

Verified step by step guidance
1
Identify the type of differential equation given: \( y'(x) = \frac{x}{y} \). This is a separable differential equation because the right side can be expressed as a function of \( x \) divided by a function of \( y \).
Rewrite the differential equation by separating variables: multiply both sides by \( y \) and multiply both sides by \( dx \) to get \( y \, dy = x \, dx \).
Integrate both sides: integrate \( y \, dy \) with respect to \( y \) and \( x \, dx \) with respect to \( x \). This gives \( \int y \, dy = \int x \, dx \).
After integrating, you will have an implicit solution involving \( y \) and \( x \) plus a constant of integration \( C \). Use the initial condition \( y(2) = 4 \) to solve for \( C \).
Finally, write the explicit solution for \( y \) in terms of \( x \) by solving the implicit equation for \( y \), if possible.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.
Video duration:
3m
Was this helpful?

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Separable Differential Equations

A separable differential equation can be written as a product of a function of x and a function of y, allowing variables to be separated on opposite sides of the equation. This enables integration with respect to each variable independently to find the general solution.
Recommended video:
06:06
Solving Separable Differential Equations

Initial Value Problems (IVP)

An initial value problem specifies a differential equation along with a condition that the solution must satisfy at a particular point. This condition helps determine the unique solution by solving for the constant of integration after finding the general solution.
Recommended video:
05:03
Initial Value Problems

Integration Techniques

Solving separable equations requires integrating both sides after separating variables. Familiarity with basic integration rules and techniques, such as power rule and substitution, is essential to find the explicit form of the solution.
Recommended video:
06:18
Integration by Parts for Definite Integrals
Related Practice
Textbook Question

Direction fields Consider the direction field for the equation y′=y(2−y) shown in the figure and initial conditions of the form y(0)=A.

d. For what values of A are the corresponding solutions decreasing, for t≥0

1
views
Textbook Question

2–10. General solutions Use the method of your choice to find the general solution of the following differential equations.

y′(t) = (2t+1)(y²+1)

1
views
Textbook Question

A predator-prey model Consider the predator-prey model

x′(t) = −4x + 2xy, y′(t) = 5y − xy

c. Find the equilibrium points for the system.

1
views
Textbook Question

Direction fields The direction field for the equation y′(t)=t−y, for |t|≤4 and |y|≤4, is shown in the figure.

b. Use the direction field to sketch the solution curve that passes through the point (0,−1/2).

1
views
Textbook Question

11–18. Solving initial value problems Use the method of your choice to find the solution of the following initial value problems.

y′(t) = -3y + 9, y(0) = 4

Textbook Question

A first-order equation Consider the equation t² y′(t) + 2ty(t) = e⁻ᵗ

c. Find the solution that satisfies the condition y(1) = 0