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

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UpStudy AI · Accepted Answer

To solve the differential equation:

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

**Step 1: Separate the Variables**

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

**Step 2: Integrate Both Sides**

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

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

**Step 3: Solve for $$ y $$**

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

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

To match the form given in the options, let $$ C' = -C $$:

$$
y(t) = \frac{t}{1 + C' t}
$$

Since $$ C $$ is an arbitrary constant, we can drop the prime notation:

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

**Equilibrium Solution:**

When $$ y(t) = 0 $$, we have an equilibrium solution because substituting $$ y = 0 $$ into the original differential equation yields $$ \frac{dy}{dt} = 0 $$.

**Final Answer:**

**Option (a)** correctly represents the general solution along with the equilibrium solution.

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

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

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