Second-Order Differential Equations Without the Function y

Second-order differential equations that don’t explicitly involve the unknown function $ y(x) $, such as $$ y'' + xy' = f(x) $$, can be solved by introducing an auxiliary variable $ u = y' $.

A Practical Example

Consider the differential equation:

$$ xy'' + y' = \log x $$

Dividing both sides by $ x $, we obtain the equation in standard form:

$$ \frac{xy'' + y'}{x} = \frac{\log x}{x} $$

$$ y'' + \frac{y'}{x} = \frac{\log x}{x} $$

Now let’s introduce the auxiliary variable $ u = y' $:

$$ y'' + \frac{u}{x} = \frac{\log x}{x} $$

Since $ u = y' $, it follows that $ y'' = u' $, yielding:

$$ u' + \frac{u}{x} = \frac{\log x}{x} $$

This is a first-order linear nonhomogeneous differential equation in $ u $.

We solve it using the integrating factor method. The standard form is $ u' + A(x)\cdot u = b(x) $, where $ A(x) = \frac{1}{x} $.

We compute the integrating factor:

$$ A(x) = e^{ \int \frac{1}{x} \ dx } = e^{\log x} = x $$

Multiplying both sides of the equation by the integrating factor $ x $:

$$ x \cdot \left( u' + \frac{u}{x} \right) = x \cdot \frac{\log x}{x} $$

$$ xu' + u = \log x $$

The left-hand side is the derivative of the product $ x \cdot u $:

$$ \frac{d}{dx}(u \cdot x) = \log x $$

We now integrate both sides with respect to $ x $:

$$ \int \frac{d}{dx}(u \cdot x) \ dx = \int \log x \ dx $$

The integral on the left is straightforward:

$$ u \cdot x = \int \log x \ dx + c_1 $$

To evaluate the integral on the right, we use integration by parts:

$$ u \cdot x = x \log x - \int x \cdot \frac{1}{x} \ dx + c_1 $$

$$ u \cdot x = x \log x - \int 1 \ dx + c_1 $$

$$ u \cdot x = x \log x - x + c_1 $$

Factoring out $ x $:

$$ u \cdot x = x ( \log x - 1 ) + c_1 $$

Since $ x > 0 $ (as it’s the argument of a logarithm), we divide both sides by $ x $:

$$ u = \log x - 1 + \frac{c_1}{x} $$

Recalling that $ u = y' $, we now integrate to recover $ y $:

$$ y' = \log x - 1 + \frac{c_1}{x} $$

Integrating both sides with respect to $ x $:

$$ \int y' \ dx = \int \left( \log x - 1 + \frac{c_1}{x} \right) dx $$

Breaking this into separate integrals using the linearity of integration:

$$ y = \int \log x \ dx - \int 1 \ dx + c_1 \int \frac{1}{x} \ dx + c_2 $$

Using the known antiderivatives:

$$ \int \log x \ dx = x(\log x - 1), \quad \int 1 \ dx = x, \quad \int \frac{1}{x} \ dx = \log x $$

Substituting these in:

$$ y = x(\log x - 1) - x + c_1 \log x + c_2 $$

Combining like terms:

$$ y = x \log x - 2x + c_1 \log x + c_2 $$

This is the general solution to the original differential equation.

And so on.