Answer:
To solve the Bernoulli equation t^2y' + 9ty − y^3 = 0, we use the substitution v = y^(1−2) = y^−1.
Differentiating v with respect to t, we get:
dv/dt = −y^−2 dy/dt
Using the chain rule, we have:
dy/dt = −y^2 dv/dt
Substituting y^−1 for v and −y^2 dv/dt for dy/dt, we get:
t^2 (−y^2 dv/dt) + 9t (1/y) y^2 − (1/v)^3 = 0
Simplifying and multiplying through by v^3, we get:
−t^2v^3 dv/dt − 9tv^2 + 1 = 0
This is now a separable differential equation. We can move the dv term to the left and the t term to the right, and then integrate both sides:
−v^−3 dv = 9t^−1 dt
Integrating both sides, we get:
v^−2/−2 = 9 ln|t| + C
Substituting v = y^−1, we get:
y^2/2 = (−1/2C) − 9 ln|t|
where C is the constant of integration.
Therefore, the solution to the given Bernoulli equation is:
y = [2/(−1/2C − 18 ln|t|)]^(1/2)
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Explanation: