Asked by Anonymous
Consider the differential equation dy/dx = x^2(y - 1). Find the particular solution to this differential equation with initial condition f(0) = 3.
I got y = e^(x^3/3) + 2.
I got y = e^(x^3/3) + 2.
Answers
Answered by
Damon
assuming that your answer is correct
dy/dx = x^2 e^(x^3/3)
but
e^(x^3/3) = y-2
so
dy/dx = x^2 (y-2) hummm
I'll try
dy/(y-1) = x^2 dx
ln(y-1)= x^3/3 + c
(y-1) = e^[x^3/3 + c]
y-1 = e^(x^3/3) e^c
y-1 = Ce^(x^3/3)
y = 1 + Ce^(x^3/3)
3 = 1 + C
C = 2
so
y = 1 + 2e^(x^3/3)
dy/dx = x^2 e^(x^3/3)
but
e^(x^3/3) = y-2
so
dy/dx = x^2 (y-2) hummm
I'll try
dy/(y-1) = x^2 dx
ln(y-1)= x^3/3 + c
(y-1) = e^[x^3/3 + c]
y-1 = e^(x^3/3) e^c
y-1 = Ce^(x^3/3)
y = 1 + Ce^(x^3/3)
3 = 1 + C
C = 2
so
y = 1 + 2e^(x^3/3)