Differential Equations Exercise 4
Consider the second-order differential equation:
$$ y'' = 6x + 2 $$
This is an elementary second-order differential equation.
To solve for the unknown function y(x), we integrate twice.
First, integrate both sides:
$$ \int y'' \, dx = \int (6x + 2) \, dx $$
The integral on the right evaluates to the antiderivative \( 3x^2 + 2x + c_1 \):
$$ \int y'' \, dx = 3x^2 + 2x + c_1 $$
The integral on the left simply gives y′:
$$ y' = 3x^2 + 2x + c_1 $$
Next, integrate once more:
$$ \int y' \, dx = \int (3x^2 + 2x + c_1) \, dx $$
The integral on the right is the antiderivative \( x^3 + x^2 + c_1x + c_2 \):
$$ \int y' \, dx = x^3 + x^2 + c_1x + c_2 $$
The integral on the left yields the unknown function y:
$$ y = x^3 + x^2 + c_1x + c_2 $$
This expression represents the general solution of the differential equation.
Here, c1 and c2 are arbitrary real constants, independent of x.
That completes the solution.
