Differential Equations Exercise 30

We want to study the following first-order differential equation:

$$ \begin{cases} y'=y^2 \cdot t^3 \\ \\ y(0)=0 \end{cases} $$

This is a classic Cauchy problem.

The equation is separable, so we can solve it by separating the variables.

First, rewrite the derivative y' using dy/dt:

$$ \frac{dy}{dt} = y^2 \cdot t^3 $$

Now separate the variables:

$$ \frac{dy}{y^2} = t^3 \, dt $$

Integrating both sides with respect to their variables gives:

$$ \int \frac{dy}{y^2} = \int t^3 \, dt $$

$$ - \frac{1}{y} + c_1 = \frac{1}{4} t^4 + c_2 $$

We can merge the constants into a single term: $c = c_2 - c_1$

$$ - \frac{1}{y} = \frac{1}{4} t^4 + c $$

$$ y = \frac{-4}{t^4 + c} $$

Next, let’s try to determine the constant using the initial condition y(0)=0, with t=0:

$$ y(0)=0 $$

$$ \frac{-4}{0 + c} = 0 $$

$$ \frac{-4}{c} = 0 $$

This is impossible, which means the general solution cannot satisfy the given Cauchy problem.

Nonetheless, by the existence theorem, a solution must exist.

The only valid solution in this case is the trivial one: y(t)=0.

$$ y(t) = 0 $$

Note. If y=0, the initial condition is satisfied: $$ y(0)= 0 $$ $$ y'=y^2 \cdot t^3 $$ $$ y'=0 \cdot t^3 $$ $$ y'=0 $$ Since the derivative of y=0 is always zero, we get y'=0, i.e. $$ 0=0 $$. The identity is satisfied.

Here, the maximal interval of existence is the entire real line.

And that’s it.