A Bernoulli differential equation is one of the form dy/dx+P(x)y=Q(x)y^n (*)

Question

A Bernoulli differential equation is one of the form dy/dx+P(x)y=Q(x)y^n (*). Observe that, if n=0 or 1 , the Bernoulli equation is linear. For other values of n , the substitution u=y^(1-n)  transforms the Bernoulli equation into the linear equation du/dx + (1-n)P(x)u=(1-n)Q(x).

Consider the initial value problem

xy'+y=3xy^2, y(1)=-3

(a) This differential equation can be written in the form (*)  with

P(x)=?

Q(x)=?

n=?

(b) The substitution u=BLANK will transform it into the linear equation du=dx+ BLANK u=BLANK

(c)Using the substitution in part (b), we rewrite the initial condition in terms of x and u

u(1)=?

(d) Now solve the linear equation in part (b), and find the solution that satisfies the initial condition in part (c).

u(x)=?

(e)(e) Finally, solve for y

y(x0=?

Answer 1

xy' + y = 3xy², y(1) = -3

This is Bernully equation: y' + (1/x)y = 3y²; Let u = y^{1 - 2} = 1/y is new function. Then u' = - y'/y². or y' = -u'/u²;

y = 1/u.

Substitute all this in our ODE: - u'/u² +(1/x)/u = 3/u², or u' - (1/x)u = -3.

Now we get linear, 1st order not homogenies ODE:

integrated factor μ = e^∫(-1/x)dx = e^-lnx = 1/x

u'·1/x - 1/x² = -3/x; (u/x)' = - 3/x; u/x = ∫(-3/x)dx; u/x = -3lnx + C; u = - 3xlnx + Cx;

y =1/u = 1/(- 3xlnx + Cx)

To get C we use initial conditions: y(1) = 3; 3 = 1/(-3·1·ln1 + C); C = 1/3; y = 3/(x - 9xlnx)