Ch. 9 - First-Order Differential Equations

Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook

Hass 15th Edition

Ch. 9 - First-Order Differential Equations

Problem 9.PE.28

Chapter 9, Problem 9.PE.28

In Exercises 23–28, solve the initial value problem.
y dx + (3x - xy + 2)dy = 0, y(2) = -1, y < 0

Verified step by step guidance

Rewrite the given differential equation in the form \(M(x,y)\,dx + N(x,y)\,dy = 0\), where \(M(x,y) = y\) and \(N(x,y) = 3x - xy + 2\).

Check if the differential equation is exact by verifying if \(\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}\). Compute \(\frac{\partial M}{\partial y}\) and \(\frac{\partial N}{\partial x}\).

If the equation is not exact, find an integrating factor that depends on either \(x\) or \(y\) to make it exact. Determine which variable the integrating factor depends on by examining the expressions for \(\frac{\partial M}{\partial y}\) and \(\frac{\partial N}{\partial x}\).

Multiply the entire differential equation by the integrating factor to obtain an exact equation. Then, find the potential function \(\Psi(x,y)\) such that \(\frac{\partial \Psi}{\partial x} = M\) and \(\frac{\partial \Psi}{\partial y} = N\).

Use the initial condition \(y(2) = -1\) to solve for the constant of integration after finding the implicit solution \(\Psi(x,y) = C\). Finally, express the solution implicitly or explicitly, considering the condition \(y < 0\).

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.

Video duration:

Was this helpful?

Key Concepts

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

Initial Value Problem (IVP)

An initial value problem involves solving a differential equation with a given initial condition, which specifies the value of the unknown function at a particular point. This condition helps determine a unique solution among many possible ones.

Exact Differential Equations

An exact differential equation can be written in the form M(x,y)dx + N(x,y)dy = 0, where there exists a function F(x,y) such that dF = M dx + N dy. Checking if ∂M/∂y = ∂N/∂x helps determine if the equation is exact, allowing integration to find the solution.

Implicit Solution and Solving for y

Often, solutions to differential equations are given implicitly as F(x,y) = C. Using the initial condition, the constant C is found. If possible, the implicit solution can be solved explicitly for y, especially considering domain restrictions like y < 0.

Recommended video:

05:14

Finding The Implicit Derivative

Related Practice

Textbook Question

In Exercises 1–22, solve the differential equation.

dy + x(2y - e^(x-x²))dx = 0

Textbook Question

In Exercises 1–22, solve the differential equation.

y' = xeʸ√(x-2)

views

Textbook Question

In Exercises 43 and 44, let S represent the pounds of salt in a tank at time t minutes. Set up a differential equation representing the given information and the rate at which S changes. Then solve for S and answer the particular questions.

Pure water flows into a tank at the rate of 4 gal/min, and the well-stirred mixture flows out of the tank at the rate of 5 gal/min. The tank initially holds 200 gal of solution containing 50 pounds of salt.

b. How many pounds of salt are in the tank after 1 minute? after 30 minutes?

views

Textbook Question

Solve the homogeneous equations in Exercises 5–10. First put the equation in the form of a homogeneous equation.

(x.exp(y/x) + y)dx - x dy = 0

views

Textbook Question

In Exercises 1–22, solve the differential equation.

x dy - (x⁴ - y) dx = 0

Textbook Question

In Exercises 1–22, solve the differential equation.

sec x dy + x cos² y dx = 0