dy/dx=x^2(y-1)

Question

dy/dx=x^2(y-1)

 

how to do this? Find the solution to the following differential equation of f(0)=3

Answer 1

y = f(x); f(0) = 3

 

dy/dx=(x^2)(y-1)

 

Separate variables and integrate:

 

∫ (1/(y-1)) dy = ∫ x^2 dx + C

 

ln|y-1| = (1/3) x^3 + C

 

y(0) = 3:

 

ln|3-1| = 0 + C

 

C = ln(2)

 

ln|y-1| = (1/3) x^3 + ln(2), y ≠ 1

 

e^(ln|y-1|) = e^((1/3) x^3 + ln(2)) 

 

|y-1| = (e^((1/3) x^3))(e^(ln(2)))

 

|y-1| = 2(e^((x^3)/3))

 

y-1 = ±2(e^((x^3)/3))

 

y = 1 ± 2(e^((x^3)/3))

 

check:

 

dy/dx = 0 ± 2( (e^((x^3)/3)) (x^2) )

 

dy/dx = (±2(e^((x^3)/3))) (x^2)

 

dy/dx = (y - 1) (x^2) √