Question

Verify that the given functions \(y_{1}\) and \(y_{2}\) satisfy the corresponding homogeneous equation; then find a particular solution of the given nonhomogeneous equation. In Problems 19 and \(20 g\) is an arbitrary continuous function. $$ (1-x) y^{\prime \prime}+x y^{\prime}-y=g(x), \quad 0

Short answer

Question: Verify if the given functions \(y_1(x) = e^x\) and \(y_2(x) = x\) satisfy the homogeneous equation \((1-x)y^{\prime\prime} + xy^{\prime} - y = 0\), and find a particular solution for the nonhomogeneous equation \((1-x)y^{\prime\prime} + xy^{\prime} - y = g(x)\). Answer: Yes, the given functions \(y_1(x) = e^x\) and \(y_2(x) = x\) satisfy the homogeneous equation \((1-x)y^{\prime\prime} + xy^{\prime} - y = 0\). The particular solution for the nonhomogeneous equation \((1-x)y^{\prime\prime} + xy^{\prime} - y = g(x)\) is \(y_p(x) = C_1 e^x + C_2 x\), where \(C_1\) and \(C_2\) are constants.

Step-by-step solution

Verify if given functions satisfy the homogeneous equation

To perform the verification, we need to substitute \(y_1(x) = e^x\) and \(y_2(x) = x\) separately into the homogeneous equation and check if it equals zero. For \(y_1(x) = e^x\): $$y_1^{\prime}(x) = e^x, \quad y_1^{\prime\prime}(x) = e^x$$ Substituting into the homogeneous equation, we get: $$(1-x)e^x + xe^x - e^x = 0 \Rightarrow e^x(1 + x - x - 1) = 0$$ Hence, \(y_1(x) = e^x\) satisfies the homogeneous equation. For \(y_2(x) = x\): $$y_2^{\prime}(x) = 1, \quad y_2^{\prime\prime}(x) = 0$$ Substituting into the homogeneous equation, we get: $$(1-x)(0) + x(1) - x = 0 \Rightarrow 0$$ Hence, \(y_2(x) = x\) also satisfies the homogeneous equation.

Find a particular solution for the nonhomogeneous equation

Let's consider \(y_p(x)\) as the particular solution for the nonhomogeneous equation. We can express \(y_p(x)\) as a linear combination of the given homogeneous solutions \(y_1(x)\) and \(y_2(x)\): $$y_p(x) = C_1 e^x + C_2 x$$ Now, we will take the first and second derivatives of \(y_p(x)\): $$y_p^{\prime}(x) = C_1 e^x + C_2, \quad y_p^{\prime\prime}(x) = C_1 e^x$$ Next, we will substitute \(y_p(x)\), \(y_p^{\prime}(x)\), and \(y_p^{\prime\prime}(x)\) into the nonhomogeneous equation: $$(1-x)(C_1 e^x) + x(C_1 e^x + C_2) - (C_1 e^x + C_2 x) = g(x)$$ We now need to obtain \(C_1\) and \(C_2\) such that the left side of the equation equals \(g(x)\). Since \(g(x)\) is an arbitrary continuous function, we will use the method of undetermined coefficients. The coefficients \(C_1\) and \(C_2\) are determined by equating the coefficients of matching powers of \(x\) on both sides of the equation: Equating the coefficients of \(x^0\) terms: $$C_1 e^x - C_1 e^x = g_0$$ Since \(C_1\) is a constant and \(0 = g_0\), there are no specific restrictions on \(C_1\). Equating the coefficients of \(x^1\) terms: $$C_2 x + C_1 x e^x - C_2 x = g_1 x$$ $$C_1 x e^x = g_1 x$$ Again, there are no specific restrictions on \(C_1\). With no restrictions on \(C_1\) and \(C_2\), the particular solution is still a linear combination of the homogeneous solutions: $$y_p(x) = C_1 e^x + C_2 x$$ The particular solution for the nonhomogeneous equation is \(y_p(x) = C_1 e^x + C_2 x\).

Key concepts

Homogeneous Equation

A homogeneous differential equation is one where every term is a function of the variable and its derivatives, without any external added functions. In our scenario, when we have the equation \( (1-x) y^{\prime \prime}+x y^{\prime}-y=0 \), with no \(g(x)\) present, it’s termed as the homogeneous part of the overall equation. This type of equation has a crucial property: if you have two solutions, say \(y_1(x)\) and \(y_2(x)\), any linear combination of these solutions is also a solution to the homogeneous equation.

It’s essential to understand that homogeneous solutions are the building blocks for solving complex differential equations, as they form the complementary part of the general solution. By verifying that \(e^x\) and \(x\) are solutions to our homogeneous equation, we reinforce the fact that these functions are intrinsically linked to the system's behavior without external influences.

Nonhomogeneous Equation

In contrast to its homogeneous counterpart, a nonhomogeneous differential equation includes an external function, in this case, the \(g(x)\). This term represents an outside force or input that disturbs the system. The equation \( (1-x) y^{\prime \prime}+x y^{\prime}-y=g(x) \) becomes nonhomogeneous due to the presence of \(g(x)\).

Solving nonhomogeneous equations typically involves finding a 'particular solution' that accounts for the nonhomogeneity and a 'complementary solution' derived from the homogeneous equation. Identifying the particular solution requires methods that often depends on the form of \(g(x)\). The significance of understanding nonhomogeneous equations lies in their ability to describe realistic systems where external factors play a role, which is common in fields like engineering and physics.

Method of Undetermined Coefficients

The method of undetermined coefficients is a technique to find the particular solution of a nonhomogeneous differential equation. It's particularly useful when the nonhomogeneous term \(g(x)\) is a polynomial, exponential, sine, or cosine function. Here’s how it works:

We assume a form for the particular solution, \(y_p(x)\), that resembles the nonhomogeneous term \(g(x)\), but with undetermined coefficients that we’ll calculate. In our case, we look for a solution in the form of a combination of \(e^x\) and \(x\), since our homogeneous solutions are of that form.

The assumed \(y_p(x) = C_1 e^x + C_2 x\) is differentiated, plugged into the nonhomogeneous equation, and the coefficients \(C_1\) and \(C_2\) are found by matching the powers of \(x\) on both sides of the equation. Although in the given problem, the coefficients are not explicitly determined due to the generic nature of \(g(x)\), this method would allow us to determine them if \(g(x)\) were a specific function.

Particular Solution

A particular solution to a differential equation is a single solution that satisfies not only the differential equation itself but also fits specific initial or boundary conditions – or in the case of a nonhomogeneous equation, incorporates the nonhomogeneity. Whatever the form of \(g(x)\), we find \(y_p(x)\) such that when substituted into the original nonhomogeneous equation, it yields \(g(x)\).

In our example, we expressed the particular solution in terms of arbitrary coefficients \(C_1\) and \(C_2\), as a linear combination of the homogeneous solutions. Its purpose is to pinpoint the response of the system to the external force described by \(g(x)\). It's critical to recognize that the general solution to the nonhomogeneous problem is the sum of this particular solution and the homogeneous (or complementary) solution.