91Ó°ÊÓ

Normalization of a differential equation is an essential initial step to simplify the equation and make it more tractable. This process involves manipulating the equation to have a leading coefficient of one, particularly important for higher-order differential equations. In our exercise, normalization is achieved by dividing the entire equation by the highest power of the dependent variable that is present with the highest derivative.

To illustrate, if we begin with the original equation \(x^{3} y^{\prime \prime}-x^{2} y^{\prime}-y=0\), dividing by \(x^3\) simplifies it to \(y^{\prime\prime}-\frac{1}{x} y^{\prime}-\frac{1}{x^{2}}y=0\). Through normalization, we can clearly see the behavior of the differential equation and further simplify it if possible, which is crucial before applying other techniques like substitution or integration.

Substitution is a method used to simplify differential equations by introducing a new variable in place of the original variables. This transformation often reveals a simpler structure that makes the equation easier to solve.

For instance, in our exercise, the substitution \(u = xy\) is used. This not only simplifies the computation but also transforms the equation into a form where standard methods of integration can be applied. We get \(u' = y + xy'\) and upon substituting into the normalized equation, the resulting form is \(u^{\prime\prime} = 0\), a much simpler second-order differential equation. This technique exemplifies how an intelligently chosen substitution can significantly reduce the complexity of a problem.

Integration is a fundamental process in solving differential equations. It's the inverse operation of differentiation and is used to find the original function, or in our case, the solution to the differential equation, given its derivative.

In the given exercise, after normalization and substitution have been applied, we are left with \(\frac{d^{2}u}{dx^{2}} = 0\). To solve this, we use integration twice: the first integration yields \(\frac{du}{dx}\), which equals a constant (since the integral of zero is a constant), and the second integration gives us \(u\) itself as a linear function of \(x\). As shown in the solution \(u = C_1 x + C_2\), where \(C_1\) and \(C_2\) are constants of integration. This step is critical as it provides the general solution to the differential equation in terms of the new variable \(u\), which we can then convert back into a solution in terms of the original variable \(y\).