y2yxy3 y02sqaureroot2SolutionSince y 2y xy3 is a Bernoulli

y\'-2y=xy^3, y(0)=2sqaureroot2

Solution

Since y\' - 2y = xy^3 is a Bernoulli DE, let y = z^(1/-(3-1)) = z^(-1/2).

Substituting this into the DE yields
(-1/2) z^(-3/2) dz/dx - 2z^(-1/2) = xz^(-3/2)
==> dz/dx + 4z = -2x.

Since this is a first order linear DE, multiply both sides by the integrating factor e^(4x):
e^(4x) dz/dx + 4e^(4x)z = -2xe^(4x)
==> (d/dx) (e^(4x)z) = -2xe^(4x), by the product rule.

Integrate both sides:
e^(4x) z = -2x * e^(4x)/4 - (-2) * e^(4x)/16 + C
==> z = (1/8 - x/2) + Ce^(-4x).

So, y = [(1/8 - x/2) + Ce^(-4x)]^(-1/2).
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Now we use the initial condition y(0) = 22:
22 = [(1/8 - 0) + C * 1]^(-1/2)
==> (22)^(-2) = 1/8 + C
==> 1/8 = 1/8 + C
==> C = 0.

So, y = (1/8 - x/2)^(-1/2)
..........= 1/(1/8 - x/2)
..........= 1/((1/8) (1 - 4x))
..........= (8)/(1 - 4x)
..........= (22)/(1 - 4x)

ans -

y = 2 sqrt(2) / sqrt(1-4x)