D.ecuation (bernoulli)?

Question

hi. how can i get to the answer

x^2 * dy/dx - 2xy = 3y^4 initial value---------->y(1)=1/2

book amswer: y^-3 = -9/5 * x^-1 + 49/5 * x^-6

Answer 1

Divide both sides by x² to put the equation in standard form:

dy/dx - (2/x)y = (3/x²)y^4

which is of the form

dy/dx + P(x) y = Q(x) y^n

The standard procedure with a Bernoulli equation is to set  

y(x) =  w(x)^(-1/(n-1))

In this case, n=4, so

y = w^(-1/3)

dy/dx = (-1/3) w^(-4/3) dw/dx

y^4 = w^(-4/3)

Inserting in the differential equation,

(-1/3) w^(-4/3) dw/dx - (2/x) w^(-1/3) = (3/x²) w^(-4/3)

Multiply both sides by -3w^(4/3):

dw/dx + (6/x) w = -9/x²

This is first order linear in w, and so has the integrating factor

e^(∫ (6/x) dx = e^(6 ln x) = e^(ln (x^6)) = x^6

Multiply both sides of the DE by x^6:

(x^6) dw/dx + (6x^5) w = -9x^4

d/dx (x^6 w) = -9x^4

Integrate both sides:

x^6 w = -(9/5)x^5 + c

w = -(9/5) x^(-1) + c x^(-6)

Going back to original variable y:

y = w^(-1/3) ⇒ w = y^(-3):

y^(-3) = -(9/5) x^(-1) + c x^(-6)

Apply initial condition:

(1/2)^(-3) = -(9/5) + c

8 = -9/5 + c

49/5 = c

Therefore,

y^(-3) = -(9/5) x^(-1) + (49/5) x^(-6)