Differential Equation Exercise
We want to solve the following differential equation:
$$ \begin{cases} y' - x^2 = 0 \\ x = 4 \\ y = \frac{1}{4} \end{cases} $$
This is a first-order differential equation, since the highest derivative involved is the first derivative y'.
It is also a Cauchy problem because two specific initial conditions are given:
$$ x=4 $$
$$ y = \frac{1}{4} $$
In other words, among the infinitely many possible solutions, we need to determine the particular solution, namely the integral curve that passes through the point (x, y) = (4, 1/4).
Let’s first rewrite the differential equation in its standard form, isolating y' as a function of the other terms:
$$ y' = x^2 $$
This is a separable differential equation of the form y' = f(x)·g(y), with f(x)=x2 and g(y)=1.
Writing the derivative in Leibniz notation gives:
$$ \frac{dy}{dx} = x^2 $$
Now separate the variables:
$$ dy = x^2 \ dx $$
Next, integrate both sides with respect to their variables:
$$ \int dy = \int x^2 \ dx $$
$$ \int 1 \ dy = \int x^2 \ dx $$
The integral on the left evaluates to F(y) = y + c1:
$$ y + c_1 = \int x^2 \ dx $$
The integral on the right is F(x) = x3/3 + c2:
$$ y + c_1 = \frac{x^3}{3} + c_2 $$
We can absorb the constants into a single constant, c = c2 - c1, where c is any real number. This gives the general solution of the equation:
$$ y = \frac{x^3}{3} + c $$
To determine the particular solution, substitute the point (x, y) = (4, 1/4) into the general solution and solve for c:
$$ y = \frac{x^3}{3} + c $$
$$ \frac{1}{4} = \frac{4^3}{3} + c $$
$$ c = \frac{1}{4} - \frac{64}{3} $$
$$ c = \frac{3-256}{12} $$
$$ c = - \frac{253}{12} $$
Therefore, the particular solution is obtained by substituting c = -253/12 into the general solution:
$$ y = \frac{x^3}{3} + c $$
$$ y = \frac{x^3}{3} - \frac{253}{12} $$
Thus, the differential equation is fully solved.
And that completes the problem.
