Domanda del libro di testo
In Exercises 1–22, solve the differential equation.
dy + x(2y - e^(x-x²))dx = 0
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Ch. 9 - First-Order Differential Equations
Hass15th EditionThomas' CalculusISBN: 9780137616077
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TOC.tableOfContents
- Ch. 1 - Functions155
- Ch. 2 - Limits and Continuity240
- Ch. 3 - Derivatives337
- Ch. 4 - Applications of Derivatives293
- Ch. 5 - Integrals41
- Ch. 6 - Applications of Definite Integrals25
- Ch. 7 - Transcendental Functions490
- Ch. 8 - Techniques of Integration399
- Ch. 9 - First-Order Differential Equations66
- Ch. 10 - Infinite Sequences and Series132
- Ch. 11 - Parametric Equations and Polar Coordinates58
Capitolo 9, Problema 9.PE.8
In Exercises 1–22, solve the differential equation.
y' = (y²-1)x⁻¹
Guida verificata passo dopo passo1
Rewrite the given differential equation in a more explicit form: \(y' = \frac{dy}{dx} = \frac{y^{2} - 1}{x}\).
Recognize that this is a separable differential equation because the right side can be expressed as a product of a function of \(y\) and a function of \(x\).
Separate the variables by rewriting the equation as \(\frac{dy}{y^{2} - 1} = \frac{dx}{x}\).
Integrate both sides: integrate \(\int \frac{1}{y^{2} - 1} \, dy\) on the left and \(\int \frac{1}{x} \, dx\) on the right.
After integration, solve the resulting equation for \(y\) implicitly or explicitly, and include the constant of integration.
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Separable Differential Equations
A separable differential equation can be written as a product of a function of y and a function of x, allowing variables to be separated on opposite sides of the equation. This form enables integration with respect to each variable independently to find the solution.
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Solving Separable Differential Equations
Integration Techniques
Solving separable equations requires integrating both sides after separation. Familiarity with integrating rational functions, such as partial fraction decomposition, is essential when dealing with expressions like (y² - 1) in the denominator or numerator.
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Percorso guidato
06:18Integration by Parts for Definite Integrals
Initial Conditions and General Solutions
After integrating, the solution typically includes an arbitrary constant representing a family of solutions. Applying initial conditions, if given, helps determine the particular solution. Understanding the difference between general and particular solutions is key.
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Percorso guidato
05:03Initial Value Problems
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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.
c. When will the tank have exactly 5 pounds of salt and how many gallons of solution will be in the tank?
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Domanda del libro di testo
In Exercises 1–22, solve the differential equation.
(x + 3y²) dy + y dx = 0 (Hint: d(xy) = y dx + x dy)
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Domanda del libro di testo
In Exercises 1–22, solve the differential equation.
y' = sin³ x cos² y
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Domanda del libro di testo
In Exercises 1–22, solve the differential equation.
2y' - y = xe^(x/2)
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Domanda del libro di testo
In Exercises 1–22, solve the differential equation.
y' = eʸ/xy
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