Question

(a) If a nontrivial solution \(\phi_{1}\) of \(L[y]=a_{0} y^{\prime \prime}+a_{1} y^{\prime}+a_{2} y=0\) is known, a second solution \(\phi_{2}\) can be obtained by varying the parameter, that is, by letting \(\phi_{2}(x)=u(x) \phi_{1}(x)\). Show that \(L[y]=g(x)\) is reduced to $$u^{\prime \prime}+\left(\frac{2 \phi_{1}^{\prime}}{\phi_{1}}+\frac{a_{1}}{a_{0}}\right) u^{\prime}=g(x)$$ which is a linear first-order equation in \(u^{\prime} .\) (b) Find a second solution of the equation $$\left(1-x^{2}\right) y^{\prime \prime}-2 x y^{\prime}+2 y=0$$ on \(0

Short answer

In summary, to find a second solution of a second-order linear homogeneous differential equation, we introduce an unknown function \(u(x)\) and use the given method to convert the equation into a first-order linear equation in \(u'(x)\). After solving this equation for \(u'(x)\), we integrate to find \(u(x)\) and compute the second solution \(\phi_2(x) = u(x) \phi_1(x)\). In this exercise, we were given the differential equation \(\left(1-x^{2}\right) y^{\prime \prime}-2 x y^{\prime}+2 y = 0\) with an initial solution of \(\phi_{1}(x) = x\). After applying the method, we obtained a first-order linear equation for \(u'(x)\) and proceeded to solve it using integration. Finally, we determine the second solution \(\phi_2(x)\) as the product of \(u(x)\) and \(\phi_1(x)\).

Step-by-step solution

Differentiation of \(\phi_{2}(x)\)

First, we need to differentiate \(\phi_2(x) = u(x)\phi_1(x)\) to get the first and second derivatives. Applying the product rule, we get: $$\phi_{2}^{\prime}(x)=u^{\prime}(x) \phi_{1}(x)+u(x) \phi_{1}^{\prime}(x)$$ And applying the product rule again to find the second derivative: $$\phi_{2}^{\prime \prime}(x) = u^{\prime \prime}(x) \phi_{1}(x) + 2u^{\prime}(x) \phi_{1}^{\prime}(x) + u(x) \phi_{1}^{\prime \prime}(x)$$

Substitute and simplify

Now we will substitute the expressions for \(\phi_2(x)\), \(\phi_2'(x)\), and \(\phi_2''(x)\) into the linear differential equation \(L[y] = a_0 y'' + a_1 y' + a_2 y = g(x)\). After substitution, it becomes: $$a_{0}\left(u^{\prime\prime}(x)\phi_{1}(x)+2 u^{\prime}(x) \phi_{1}^{\prime}(x)+u(x) \phi_{1}^{\prime \prime}(x)\right)+ a_{1}\left(u^{\prime}(x) \phi_{1}(x)+u(x) \phi_{1}^{\prime}(x)\right) + a_{2}\left(u(x) \phi_{1}(x)\right) = g(x)$$ Now, since \(\phi_{1}(x)\) is already a solution, we know that \(L[\phi_{1}] = a_{0}\phi_{1}^{\prime\prime}(x) + a_{1}\phi_{1}^{\prime}(x) + a_{2}\phi_{1}(x) = 0\). Therefore, we can simplify the above equation to get: $$a_{0}\left(u^{\prime\prime}(x)\phi_{1}(x)+2 u^{\prime}(x) \phi_{1}^{\prime}(x)\right)+ a_{1}\left(u^{\prime}(x) \phi_{1}(x)\right) = g(x)$$

Rearrange into first-order linear equation for \(u'(x)\)

Dividing both sides by \(a_0\phi_{1}(x)\), we have: $$u^{\prime\prime}(x) + 2\frac{\phi_{1}^{\prime}(x)}{\phi_{1}(x)}u^{\prime}(x) + \frac{a_{1}}{a_0} u^{\prime}(x) = \frac{g(x)}{a_0\phi_{1}(x)}$$ Now let \(k(x) = 2\frac{\phi_{1}^{\prime}(x)}{\phi_{1}(x)} + \frac{a_{1}}{a_0}\). Then, we can rewrite the equation as: $$u^{\prime\prime}(x) + k(x) u^{\prime}(x) = \frac{g(x)}{a_0\phi_{1}(x)}$$ This is a linear first-order equation in \(u'(x)\).

Part (b) Method application

Given the equation \(\left(1-x^{2}\right) y^{\prime \prime}-2 x y^{\prime}+2 y = 0\), we are asked to find a second solution \(\phi_{2}(x)\) with the initial solution being \(\phi_{1}(x)=x\). We can use the result obtained in part (a) to rewrite the given equation as a first-order linear equation for \(u'(x)\): $$u^{\prime\prime}(x) + k(x) u^{\prime}(x) = 0$$ where \(k(x) = 2\frac{x'}{x} - \frac{2x}{1-x^2} = 2 - \frac{2x}{1-x^2}\).

Solve the equation for \(u'(x)\)

To solve the first-order linear equation, we can use an integrating factor. The integrating factor is \(\exp{\int k(x) dx} = \exp{\int (2 - \frac{2x}{1-x^2}) dx}\). After integration, we have: $$u^{\prime}(x) e^{\int { (2 - \frac{2x}{1-x^2}) dx }} = c_1$$ By integrating once more, we will find the function \(u(x)\). Then we can compute the second solution: \(\phi_2(x) = u(x) \phi_1(x)\).

Key concepts

Linear Differential Equation

Linear differential equations are a fundamental class of differential equations important in engineering and physics. Such an equation is of the form:

• \[ L[y] = a_0(x)y'' + a_1(x)y' + a_2(x)y = g(x) \]

For a linear differential equation, the function \( y \) and its derivatives appear linearly. This means there are no products or powers of \( y, y', \) or \( y'' \).
In our exercise, the goal was to work with a homogeneous linear differential equation where \( g(x) = 0 \). Once you have a solution \( \phi_1(x) \), you can find a second solution using variation of parameters.
Understanding linearity helps as it allows us to superpose (add) solutions, which is crucial in solving complex problems in real-world applications.

Second Solution

Finding a second solution when one solution is already known involves a technique called "variation of parameters." If \( \phi_1(x) \) is a known solution, we look for the second solution in the form:

• \[ \phi_2(x) = u(x)\phi_1(x) \]

Here \( u(x) \) is a function to be determined.
To find \( \phi_2(x) \), we differentiate \( u(x)\phi_1(x) \) using the product rule. Substitution back into the original equation helps in determining the form of \( u(x) \). This leads us to a simpler first-order differential equation for \( u'(x) \).
This method is particularly useful when dealing with complex differential equations or systems that allow for analytical solutions, enabling the discovery of linearly independent solutions.

Nontrivial Solution

A nontrivial solution of a differential equation is any solution that is not the identically zero solution. In the context of our exercise:

• If \( \phi_1(x) \) is nontrivial, it means \( \phi_1(x) eq 0 \) for all \( x \) in the domain.

Nontrivial solutions are essential in constructing the general solution of differential equations. They allow us to span the solution space fully.
When solving a differential equation, especially one that is homogeneous, identifying nontrivial solutions helps ensure the uniqueness of the solution set. Without them, you cannot form the basis required to express other potential solutions. This is closely linked with the use of the Wronskian in determining linear independence between solutions.