2. (2 points) Find the general solution of the following differential equation: $$ \frac{d y}{d t}=\frac{y^{2}}{t^{2}} \quad \frac{d y}{y^{2}}=\frac{d t}{t^{2}} $$ (a) $$ y(t)=\frac{t}{1+C t} $$, together with equilibrium solution $$ y(t)=0-\frac{d}{y^{2}}= $$ (b) $$ y(t)=t+C $$, together with equilibrium solution $$ y(t)=0 $$ (k) $$ y(t)=\frac{-t}{1+C t} $$, together with equilibrium solution $$ y(t)=0 $$ (d) $$ y(t)=C t $$

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To solve the differential equation:

$$
\frac{dy}{dt} = \frac{y^{2}}{t^{2}}
$$

we can separate the variables:

$$
\frac{dy}{y^{2}} = \frac{dt}{t^{2}}
$$

Integrating both sides:

$$
\int \frac{dy}{y^{2}} = \int \frac{dt}{t^{2}}
$$

$$
- \frac{1}{y} = - \frac{1}{t} + C
$$

Solving for $$ y $$:

$$
\frac{1}{y} = \frac{1}{t} - C
$$

$$
y = \frac{1}{\frac{1}{t} - C} = \frac{t}{1 - C t}
$$

To match the given options, we can rewrite the constant $$ C $$ as $$ -C' $$, resulting in:

$$
y(t) = \frac{t}{1 + C t}
$$

Additionally, the equilibrium solution occurs when $$ \frac{dy}{dt} = 0 $$, which implies $$ y(t) = 0 $$.

**Answer:**  
**Option (a)**: $$ y(t) = \dfrac{t}{1 + C t} $$, together with equilibrium solution $$ y(t) = 0 $$
The general solution is $$ y(t) = \frac{t}{1 + C t} $$, and the equilibrium solution is $$ y(t) = 0 $$.

(2 points) Find the general solution of the following differential equation:

(a) , together with equilibrium solution
(b) , together with equilibrium solution
(k) , together with equilibrium solution
(d)

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(2 points) Find the general solution of the following differential equation: (a) , together with equilibrium solution (b) , together with equilibrium solution (k) , together with equilibrium solution (d)