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Ch. 9 - Differential Equations
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Non è quello che usi tu?Cambia libro di testo
Capitolo 9, Problema 9.R.9

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)

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1
Recognize that the given differential equation is separable since it can be written as \(y'(t) = (2t+1)(y^2 + 1)\), which allows us to separate variables involving \(y\) and \(t\) on opposite sides.
Rewrite the equation in differential form as \(\frac{dy}{dt} = (2t+1)(y^2 + 1)\), then separate variables to get \(\frac{dy}{y^2 + 1} = (2t + 1) dt\).
Integrate both sides: integrate \(\int \frac{dy}{y^2 + 1}\) with respect to \(y\) and \(\int (2t + 1) dt\) with respect to \(t\).
Recall that \(\int \frac{dy}{y^2 + 1} = \arctan(y) + C_1\) and \(\int (2t + 1) dt = t^2 + t + C_2\), where \(C_1\) and \(C_2\) are constants of integration.
Combine the results to write the implicit general solution as \(\arctan(y) = t^2 + t + C\), where \(C\) is a constant that absorbs \(C_1\) and \(C_2\). Optionally, solve for \(y\) by taking the tangent of both sides to express \(y\) explicitly.

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