Differential Equations Exercise 29
Consider the first-order differential equation:
$$ \begin{cases} y'=y^2 \cdot x^3 \\ \\ y(0)=7 \end{cases} $$
This is a Cauchy problem involving a separable differential equation of the form y’ = f(x)g(y), with f(x) = x^3 and g(y) = y^2.
Rewriting in Leibniz notation gives:
$$ \frac{dy}{dx}=y^2 \cdot x^3 $$
Separating the variables yields:
$$ \frac{dy}{y^2}=x^3 \ dx $$
Integrating both sides, we obtain:
$$ \int \frac{1}{y^2} \, dy = \int x^3 \, dx $$
Carrying out the integrations leads to:
$$ -\frac{1}{y} + c_1 = \frac{1}{4} x^4 + c_2 $$
By combining the constants into a single term, c = c2 − c1, we have:
$$ -\frac{1}{y} = \frac{1}{4} x^4 + c $$
Solving for y gives:
$$ y = \frac{-1}{\tfrac{1}{4} x^4 + c} $$
$$ y = \frac{-1}{\tfrac{x^4 + 4c}{4}} $$
$$ y = \frac{-4}{x^4 + 4c} $$
Since 4c is itself a constant, we may simply denote it again by c. Thus, the general solution is:
$$ y = \frac{-4}{x^4 + c} $$
To determine the particular solution, apply the initial condition y(0) = 7:
$$ 7 = \frac{-4}{0^4 + c} $$
$$ 7 = \frac{-4}{c} $$
Hence,
$$ c = \frac{-4}{7} $$
Substituting this value into the general solution gives:
$$ y = \frac{-4}{x^4 + c} $$
$$ y = \frac{-4}{x^4 - \tfrac{4}{7}} $$
$$ y = \frac{-4}{\tfrac{7x^4 - 4}{7}} $$
$$ y = \frac{-28}{7x^4 - 4} $$
Therefore, the solution to the Cauchy problem is:
$$ y = \frac{-28}{7x^4 - 4} $$
This completes the solution.
