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Ch. 7 - Transcendental Functions
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Non è quello che usi tu?Cambia libro di testo
Capitolo 7, Problema 7.4.10

Solve the differential equation in Exercises 9–22.
10. (dy/dx) = x²√y, y > 0

Guida verificata passo dopo passo
1
Recognize that the given differential equation is separable. The equation is \(\frac{dy}{dx} = x^{2} \sqrt{y}\), where \(y > 0\).
Rewrite the equation to separate variables \(y\) and \(x\). Express it as \(\frac{dy}{\sqrt{y}} = x^{2} dx\).
Integrate both sides: integrate \(\frac{1}{\sqrt{y}} dy\) on the left and $x^{2} dx$ on the right.
After integrating, you will get an expression involving \(y\) and \(x\) plus an integration constant \(C\). Write this implicit solution.
Finally, solve the implicit equation for \(y\) if possible, to express \(y\) explicitly as a function of \(x\).

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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 solution.
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Integration of Power Functions

Integrating power functions involves applying the power rule for integration, which states that ∫x^n dx = (x^(n+1))/(n+1) + C for n ≠ -1. This is essential when integrating terms like x² or y raised to a fractional power.
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Initial Conditions and Domain Restrictions

The condition y > 0 restricts the solution to positive values of y, which affects the domain and the form of the solution. Understanding initial or domain conditions ensures the solution is valid and meaningful within the given constraints.
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