Question

Use the method of Problem 23 to solve the given equation for \(x>0 .\) \(x^{2} y^{\prime \prime}-2 x y^{\prime}+2 y=3 x^{2}+2 \ln x\)

Short answer

Question: Determine the general solution of the given second-order non-homogeneous differential equation: \(x^2y'' - 2xy' + 2y = 3x^2 + 2\ln{x}\) for x > 0. Answer: The general solution of the given differential equation is \(y = x^{-2}(Ce^{-x} - 12x^2 e ^{-\frac{1}{2}x} + 24xe^{-\frac{1}{2}x} - 48e^{-\frac{1}{2}x} - 4\ln{x})\), where C is an arbitrary constant.

Step-by-step solution

Determine the Transforming Function

We first need to find the transforming function, which is \(e^{\int R(x) dx}\), where R(x) is the coefficient of \(y'\) divided by the coefficient of \(y''\). In this case, R(x) = \(\frac{-2x}{x^2}\) = \(-\frac{2}{x}\) Now we can find the transforming function: \(e^{\int R(x) dx}=e^{\int -\frac{2}{x} dx}\) = \(e^{-2\ln{x}}\) = \(x^{-2}\)

Substitute y

Substituting \(y = x^{-2} v(x)\) and its derivatives, \(y' = -2x^{-3}v + x^{-2}v'\) and \(y''=-6x^{-4}vang-4x^{-3}vang\) into the original equation: \(x^2 (-6x^{-4}v - 4x^{-3}v') - 2x(-2x^{-3}v + x^{-2}v') + 2(x^{-2}v)=3 x^{2}+2 \ln x\)

Simplify Equation and Separate Homogeneous and Non-Homogeneous parts

Simplify the above equation: \(-6v - 4xv' + 4v - 2xv' = 3x^2 + 2\ln{x}\) Combine the two parts: \(-2v' = 3x^2 + 2\ln{x} + 2v\) Now, we have transformed the second-order differential equation into two separate first-order non-homogeneous linear differential equations. One for the homogeneous equation: \(-2v'_h = 2v_h\) And one for the particular solution: \(-2v'_p = 3x^2 + 2\ln{x}\)

Solve Homogeneous Equation

We can solve the homogeneous equation: \(-2v'_h = 2v_h \Rightarrow v'_h = -v_h\) This equation has the general solution: \(v_h = Ce^{-x}\), where C is an arbitrary constant.

Solve Non-Homogeneous Equation

The non-homogeneous equation is: \(-2v'_p = 3x^2 + 2\ln{x}\) We will use an integrating factor, which will be given by: \(e^{\int\frac{1}{-2}dx} = e^{-\frac{1}{2}x}\) Multiplying the equation by the integrating factor: \(-2e^{-\frac{1}{2}x}v'_p = (3x^2 + 2\ln{x})e^{-\frac{1}{2}x}\) Now, we can write the left side as a derivative of a product: \(\frac{d}{dx}\left(-2v_pe^{-\frac{1}{2}x}\right) = (3x^2 + 2\ln{x})e^{-\frac{1}{2}x}\) Integrate both sides with respect to x: \(v_p = -\int (3x^2 + 2\ln{x})e^{-\frac{1}{2}x} dx\) To solve this integral, we use integration by parts. The final result will be: \(v_p=-12x^2e^{-\frac{1}{2}x} + 24xe^{-\frac{1}{2}x} - 48e^{-\frac{1}{2}x} - 4 \ln{x}\)

Combine the Homogeneous and Non-Homogeneous Solutions

Now we can find the general solution for the non-homogeneous differential equation as the sum of the homogeneous and non-homogeneous solutions: \(y = x^{-2}(v_h + v_p) = x^{-2}(Ce^{-x} - 12x^2 e ^{-\frac{1}{2}x} + 24xe^{-\frac{1}{2}x} - 48e^{-\frac{1}{2}x} - 4\ln{x})\) This is the general solution of the original differential equation.

Key concepts

Second Order Differential Equations

Understanding second order differential equations is foundational in studying various physical and engineering systems. These equations involve derivatives of at least the second order, which often relates to acceleration in mechanical contexts or curvature in geometrical ones.

For instance, a general second-order differential equation takes the form: \[y'' + p(x)y' + q(x)y = g(x)\] where \(y''\) represents the second derivative of \(y\), \(p(x)\) and \(q(x)\) are coefficients that can depend on \(x\), and \(g(x)\) is a known function. If \(g(x) = 0\), the equation is termed homogeneous; otherwise, it's non-homogeneous. Such equations frequently appear in contexts such as oscillations, electrical circuits, and fluid dynamics. In our exercise, we manipulated a complicated second-order equation into a more manageable format to find a solution, showcasing the versatility of techniques available for solving these equations.

Integrating Factor Method

The integrating factor method is a potent tool for tackling first-order linear differential equations. When we have an equation of the form \(y' + P(x)y = Q(x)\), we seek an integrating factor, typically denoted \(\mu(x)\), that once multiplied by all terms, transforms the left side into the derivative of a product. This factor is conventionally expressed as \(\mu(x) = e^{\int P(x)dx}\).

By applying this approach, as in Step 5 of the original solution, the differential equation becomes more tractable. The integration process then often simplifies to computing the integral of the right side of the transformed equation. This method was crucial in our step-by-step example to finding the particular solution for the non-homogeneous part.

Homogeneous and Non-Homogeneous Equations

Homogeneous differential equations are those where the function \(g(x)\) is zero, leading to the form \[y'' + p(x)y' + q(x)y = 0\]. These equations, given certain conditions, have solutions that can be superposed, which is a valuable property for constructing general solutions. In contrast, non-homogeneous differential equations like \[y'' + p(x)y' + q(x)y = g(x)\] (with \(g(x) \eq 0\)) represent the presence of an external forcing or input.

In our example, step 3 involved simplifying the original equation to separate the homogeneous and non-homogeneous components. This enabled solving them independently, where the homogeneous solution captures the structure of the system's response, and the non-homogeneous solution accounts for the forced response due to an external factor.

Integration by Parts

Integration by parts is a fundamental technique in calculus, encapsulating the product rule of differentiation for integrals. It can be summarized as: \[\int u dv = uv - \int v du\], where \(u\) and \(dv\) are chosen from the integrand parts. The selection of \(u\) and \(dv\) can greatly influence the simplicity of the subsequent integral.

In the provided solution, step 5 calls for integration by parts to solve for the particular integral. This technique is often essential when the integrand is the product of two different classes of functions, as we saw with \(x^2\) and the exponential function. Mastering this method not only helps solve differential equations but also supports a broad range of applications in both pure and applied mathematics.

Differential Equations Step by Step Solution

Providing a step-by-step solution to differential equations enables a structured approach to breaking down complex problems into manageable parts. This pedagogical strategy aids students in understanding every aspect of the problem-solving process. As in our example, we began by finding an appropriate transforming function, followed by substituting variable terms to simplify the equation, then separating the equation into homogeneous and non-homogeneous parts before solving them individually.

Each step of the process sheds light on a different mathematical principle or technique, such as the integrating factor method or integration by parts, leading students to the final solution. These incremental insights are invaluable not just for this particular problem but also build a conceptual framework applicable to a variety of differential equations. Clear step-by-step solutions contribute to deeper comprehension and can significantly enhance problem-solving skills among learners.