Solving the Initial Value Problem for x'=x^3 with x(0)=1

[simo1]

solve the initial value problem:
x'=x^3 x(1)=1

my work

dx/x^3 =dt
then I integrated wrt t and obtained
x^(-2) = t + c(c0nstant)
where then
this is 1/x^2 =t+c
1/x = square root of (t+c)
then
x= 1/sqrt(t+c)

now when i apply the Initial value problem i get c = 0 and that is incorrect. where am i going wrong

 

[onie mti]

the intergral of x^-3 is (1/2)multiply by 1/x^2=t+c
then your constant value will be 3:)

 

[Prove It]

the intergral of x^-3 is (1/2)multiply by 1/x^2=t+c
then your constant value will be 3:)

Actually it's -(1/2)(1/x^2), or -(1/2)x^{-2}