Differential Equations - Exercise 5

Let’s look at the differential equation:

$$ xy'' = 1 $$

This is a second-order differential equation, since the highest derivative involved is the second derivative $y''$.

We can first solve for $y''$ explicitly:

$$ y'' = \frac{1}{x} $$

To obtain the general solution, we integrate twice.

First, integrate both sides:

$$ \int y'' \, dx = \int \frac{1}{x} \, dx $$

The right-hand side integrates to $\log(x) + c_1$, so:

$$ y' = \log(x) + c_1 $$

Next, we integrate once again:

$$ \int y' \, dx = \int \big(\log(x) + c_1\big) \, dx $$

The integral evaluates to $x \log(x) - x + c_1x + c_2$.

Therefore,

$$ y = x \log(x) - x + c_1x + c_2 $$

This gives the general solution of the differential equation.

And so on.