Show that $\frac{−1}{t}$ is a solution of $y′=y^2$ passing through $(−1, 1)$.

Question

I have to prove the following statement:

Show that $\dfrac{−1}{t}$ is a solution of $y′=\dfrac{dy}{dt}=y^2$ passing through $(−1, 1)$.

I'm not sure where to start. If anyone could help me go step by step, I'd really appreciate it.

Answer 1

To show that

$y(t) = -\dfrac{1}{t} \tag 1$

is a solution to

$y' = y^2, \tag 2$

all we need do is verify that $y(t)$ as in (1) obeys (2). From (1),

$y'(t) = \left ( -\dfrac{1}{t} \right )' = (-t^{-1} )' = t^{-2}; \tag 3$

now we also have

$(y(t))^2 = \left (-\dfrac{1}{t} \right )^2 = \dfrac{1}{t^2} = t^{-2}; \tag 4$

comparing (3) and (4) shows that (1) is a solution of (2); to show it passes through $(-1, 1)$ is equally simple:

$y(-1) = -\dfrac{1}{-1} = 1; \tag 5$

since $y(-1) = 1$, the solution contains the point $(-1, 1)$.