A Bernoulli differential equation is one of the form dydxPxy

Solution

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A Bernoulli differential equation is on of the form

dy/dx + P(x)y = Q(x)yⁿ

Observe that, if n = 0 or 1, the Bernoulli equation is linear. For the other values of n, the substitution y = y^1-n transforms the Bernoulli equation into the linear equation

du/dx + (1-n)P(x)u = (1-n)Q(x).

Use an appropriate substitution to solve the equation

xy\' + y = -6xy^2,

and find the solution that satisfies y(1) = -1.

Answer

xy\' + y = -6xy^2

Divide each term by x

y\' + (1/x)y = -6y^2

Now divide by y^2

y\'/y^2 + 1/(xy) = -6

Let u = 1/y so u\' = du/dx = du/dy dy/dx = -1/y^2*dy/dx = -y\'/y^2

Substituting we have

-u\' + (1/x)u = -6
u\' - (1/x)u = 6

This is a linear differential equation.

Let v = 1/x so v\' = -1/x^2 so v\'/v = -1/x

Now multiply by the integrating factor v

u\'v - (1/x)uv = 6v
u\'v + (v\'/v)uv = 6v
u\'v + v\'u = 6v

The LHS is just the product rule applied to uv - integrating we get

uv = 6ln(x) + C <--- remember v = 1/x

u = 6xln(x) + Cx

1/y = x[6ln(x) + C]

y = 1/{x[6ln(x) + C]}
When x = 1 y = -1
c=1