Why the ODE x′(t)=x2(t) doesn't have solutions if x(0)=x0≠0

x′(t)=x2(t)dxdt=x2dxx2=dt∫dxx2=∫dt−1x=t+cx(t)=−1t+cif x(0)=x0 thenx(0)=−10+c=−1c=x0(x0≠0)c=−1x0x(t)=−1t+(−1x0)=x01−x0tLet's find the values that x(t) is not defined:1−x0t=0t=1x0if t=1x0 then x(t) is not defined. x(t) is defined for all other values except t=1x0(x0≠0).

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maduregimc

Answered question

2022-01-22

Why the ODE

x^{'} (t) = x^{2} (t)

doesn't have solutions if

x (0) = x_{0} \neq 0

Answer & Explanation

trisanualb6

Beginner2022-01-22Added 32 answers

x^{'} (t) = x^{2} (t)

\frac{d x}{d t} = x^{2}

\frac{d x}{x^{2}} = d t

\int \frac{d x}{x^{2}} = \int d t

\frac{- 1}{x} = t + c

x (t) = \frac{- 1}{t + c}

x (0) = x_{0}

then

x (0) = \frac{- 1}{0 + c} = \frac{- 1}{c} = x_{0} (x_{0} \neq 0)

c = \frac{- 1}{x_{0}}

x (t) = \frac{- 1}{t + (\frac{- 1}{x_{0}})} = \frac{x_{0}}{1 - x_{0} t}

Let's find the values that x(t) is not defined:

1 - x_{0} t = 0

t = \frac{1}{x_{0}}

t = \frac{1}{x_{0}}

then x(t) is not defined. x(t) is defined for all other values except

t = \frac{1}{x_{0}} (x_{0} \neq 0)

chumants6g

Beginner2022-01-23Added 33 answers

Nobody has said that any initial value problem you cook up will have a solution for all t - in the same vein that you cannot expect a real solution for all equations of the form

x^{2} = c \in R

.
On the positive side you can say the following: When the hypotheses of the existence and uniqueness theorem are satisfied throughout the domain

Ω \subset R^{2}

of the given ODE

x^{'} = f (x, t)

then the graph of any ''complete'' solution will finally leave any given compact

K \subset Ω

.
In the case at hand the graph of the solution

y (t) = \frac{x_{0}}{1 - t x_{0}}

will go off to

\infty

when

t x_{0} \to 1 -

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