Give all solutions x(t) of the differential equation x-dot = 4t x^2.

Step-by-step solution

1. 1

   Spot that the equation separates. The right-hand side factors as (a function of $t$) times (a function of $x$): $4t\cdot x^2$. That is exactly the condition for separation of variables.

2. 2

   Separate and integrate. Assuming $x\ne 0$ so that dividing is legal, $\frac{dx}{x^2}=4t\,dt\ \Longrightarrow\ \int x^{-2}\,dx=\int 4t\,dt\ \Longrightarrow\ -\frac1x=2t^2+C_0.$

3. 3

   Solve for $x$. Multiplying by $-1$ gives $\dfrac1x=-2t^2-C_0$. Writing $C=-C_0$ absorbs the sign: $x(t)=\frac{1}{C-2t^2},\qquad C\in\mathbb{R}.$ Each such solution lives only on an interval where $C-2t^2\ne 0$; for $C>0$ the line blows up at $t=\pm\sqrt{C/2}$.

4. 4

   Recover the solution that separation threw away. Dividing by $x^2$ silently assumed $x\ne 0$. Test $x(t)\equiv 0$ directly: $\dot{x}=0$ and $4tx^2=0$, so it satisfies the equation. It is a genuine solution and it is not of the form $1/(C-2t^2)$ for any finite $C$.

5. 5

   Verify the family. Differentiating $x=(C-2t^2)^{-1}$ gives $\dot{x}=-(C-2t^2)^{-2}(-4t)=\dfrac{4t}{(C-2t^2)^2}=4tx^2$, as required.

6. 6

   State the complete solution set. $x(t)\equiv 0\qquad\text{or}\qquad x(t)=\frac{1}{C-2t^2}\ \ (C\in\mathbb{R}),$ each on a maximal interval where the denominator stays away from zero.