Q. The initial value problem is \(x'(t) = (t-1) x^2(t)\), \(x(0) = -8\). Find \(x(1)\).

Q: How do I solve the ODE x'(t)=(t-1)x^2(t) with x(0)=-8?

Separate variables: \mathrm{d}x/x^2=(t-1)\,\mathrm{d}t . Integrate: -1/x = t^2/2 - t + C . Use x(0)=-8 to get C=1/8 . So x(t)=-1/\left(t^2/2 - t + 1/8\right) on intervals where the denominator is nonzero..

Q: Where does the solution blow up (finite-time singularity)?

Blow-up occurs where t^2/2 - t + 1/8 = 0. Solving gives t = 1 \pm \sqrt3/2. The first singularity after 0 is t_* = 1 - \sqrt3/2 \approx 0.133975..

Q: Is the solution unique?.

Yes. f(t,x)=(t-1)x^2 is continuous and locally Lipschitz in x, so Picard–Lindelöf guarantees a unique local solution through (0,-8) up to the maximal existence time (the blow-up).

Q: What is the maximal interval of existence containing t=0?.

The maximal interval containing 0 is (-\infty,\;1-\sqrt3/2). The solution cannot be continued past t_*=1-\sqrt3/2 because |x(t)|\to\infty there.

Q: What is the local behavior of x(t) near the blow-up time?.

Near t_*, the denominator is approximately linear, so x(t)\sim -1/(C'(t-t_*)). Thus |x(t)| diverges like a simple pole as t\to t_*^\mp..

Answer

Separate variables: \(\frac{dx}{x^2}=(t-1)\,dt\). Integrate: \(\int x^{-2}dx=\int (t-1)dt\) gives \(-\frac{1}{x}=\frac{t^2}{2}-t+C\). With \(x(0)=-8\) we get \(C=\frac{1}{8}\). Hence \(\displaystyle -\frac{1}{x}=\frac{t^2}{2}-t+\frac{1}{8}\), so \(x(t)=\frac{8}{-4t^2+8t-1}\). Therefore \(x(1)=\frac{8}{3}\).

Detailed Explanation

Solution (step-by-step)

Write the differential equation and initial condition:
\(x'(t)= (t-1)\,x^2(t)\), \(\; x(0)=-8\).
Recognize this is a separable equation. Separate variables so that all x-terms are on one side and t-terms on the other:
\(\dfrac{dx}{x^2} = (t-1)\,dt\).
Integrate both sides with respect to their variables:
\(\displaystyle \int x^{-2}\,dx = \int (t-1)\,dt.\)

Compute each integral:

\(\displaystyle -\,\frac{1}{x} = \frac{1}{2}t^2 – t + C,\) where \(C\) is an arbitrary constant of integration.
Use the initial condition \(x(0)=-8\) to determine \(C\). Substitute \(t=0\) and \(x=-8\) into the integrated equation:
\(\displaystyle -\frac{1}{-8} = \frac{1}{2}\cdot 0^2 – 0 + C.\)

Thus \(\displaystyle \frac{1}{8} = C.\)
Substitute \(C=\tfrac{1}{8}\) back into the relation and solve for \(x(t)\):
\(\displaystyle -\frac{1}{x} = \frac{1}{2}t^2 – t + \frac{1}{8}.\)

Therefore

\(\displaystyle x(t) = -\frac{1}{\tfrac{1}{2}t^2 – t + \tfrac{1}{8}}.\)

Optionally simplify by multiplying numerator and denominator by 8:

\(\displaystyle x(t) = -\frac{8}{4t^2 – 8t + 1}.\)
Evaluate at \(t=1\):
\(\displaystyle x(1) = -\frac{8}{4\cdot 1^2 – 8\cdot 1 + 1} = -\frac{8}{4 – 8 + 1} = -\frac{8}{-3} = \frac{8}{3}.\)

Answer: \(x(1)=\dfrac{8}{3}\).

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Calculus FAQs

How do I solve the ODE \(x'(t)=(t-1)x^2(t)\) with \(x(0)=-8\)?

Separate variables: \( \mathrm{d}x/x^2=(t-1)\,\mathrm{d}t \). Integrate: \( -1/x = t^2/2 - t + C \). Use \( x(0)=-8 \) to get \( C=1/8 \). So \( x(t)=-1/\left(t^2/2 - t + 1/8\right) \) on intervals where the denominator is nonzero..

What is \(x(1)\)?

The initial-value solution blows up before t=1, so x(1) is not defined for that solution. Formally plugging t=1 into the algebraic expression gives \(8/3\), but that value is not reachable from the IVP because of a finite-time singularity.

Where does the solution blow up (finite-time singularity)?

Blow-up occurs where \(t^2/2 - t + 1/8 = 0\). Solving gives \(t = 1 \pm \sqrt3/2\). The first singularity after 0 is \(t_* = 1 - \sqrt3/2 \approx 0.133975\)..

Is the solution unique?.

Yes. \(f(t,x)=(t-1)x^2\) is continuous and locally Lipschitz in \(x\), so Picard–Lindelöf guarantees a unique local solution through \((0,-8)\) up to the maximal existence time (the blow-up).

What is the maximal interval of existence containing \(t=0\)?.

The maximal interval containing 0 is \((-\infty,\;1-\sqrt3/2)\). The solution cannot be continued past \(t_*=1-\sqrt3/2\) because \(|x(t)|\to\infty\) there.

Can the solution be extended past the singularity by choosing a different branch?

Why does plugging \(t = 1\) into the closed-form solution give a misleading result?

The algebraic formula is valid where the solution branch from the initial condition is defined. Because the branch has a vertical asymptote at \(t_*\approx 0.134\), values at \(t=1\) are not part of that branch; substituting ignores the domain restriction.

What is the local behavior of \( x(t) \) near the blow-up time?.

Near \(t_*\), the denominator is approximately linear, so \(x(t)\sim -1/(C'(t-t_*))\). Thus \(|x(t)|\) diverges like a simple pole as \(t\to t_*^\mp\)..

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