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Ch. 9 - First-Order Differential Equations
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Non è quello che usi tu?Cambia libro di testo
Capitolo 9, Problema 9.PE.21

In Exercises 1–22, solve the differential equation.


y' = xy ln x ln y

Guida verificata passo dopo passo
1
Rewrite the given differential equation \(y' = xy \ln x \ln y\) in Leibniz notation as \(\frac{dy}{dx} = xy \ln x \ln y\).
Separate the variables by dividing both sides to isolate \(y\) terms on one side and \(x\) terms on the other: write it as \(\frac{1}{y \ln y} dy = x \ln x \, dx\).
Integrate both sides separately: compute \(\int \frac{1}{y \ln y} dy\) on the left and \(\int x \ln x \, dx\) on the right.
For the left integral, use substitution \(u = \ln y\) which implies \(du = \frac{1}{y} dy\), transforming the integral into \(\int \frac{1}{u} du\).
For the right integral, use integration by parts where you let \(u = \ln x\) and \(dv = x \, dx\), then apply the formula \(\int u \, dv = uv - \int v \, du\).

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