Question

Solve the general solution of the equation xydx +2(x² + 2y²)dy = 0​

Answer 1

Answer:

3x² + 2y³ = C

Step-by-step explanation:

(x² + 2y²) = -2xydx

dx/dy = x² + 2y²/-2xy

Let x = vy, dx/dy = v + y • dv/dy

v + y • dv/dy = v²y² + 2y²/-2vy² = y²(v² + 2)/-2vy²

ydv/dy = v² + 2/-2v - v = v² + 2 + 2v²/-2v

ydv/dy = 3v² + 2/-2v

Integrate both sides, we get

3v² + 2/-2v (dv) = dy/y

1/3 dt/t = -dy/y

1/3 logt = -logy +logc

logt 1/3 = log(c • y)

logt⅓ = c/y

==> t = c³/y³

3v² + 2 = c³/y³

y³ = (3x² + 2)/y² = c³

==> 3x² + 2y³ = c

Answer 2

Answer:

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