Question

Use the method of Problem 23 to solve the given equation for \(x>0 .\) \(x^{2} y^{\prime \prime}+7 x y^{\prime}+5 y=x\)

Short answer

#Answer# The general solution of the given differential equation is: $$y(x) = C_1x^{-1} + C_2x^{-5} + \frac{1}{5}x$$

Step-by-step solution

Identifying the homogeneous equation and complementary function

First, let's identify the homogeneous equation of the given equation. The homogeneous equation will have the same left-hand side as the given equation but its right-hand side will be replaced by 0: $$x^{2}y'' + 7xy' + 5y = 0$$ To find the complementary function of this homogeneous equation, we first replace \(y'\) with \(\frac{dy}{dx}\) and \(y''\) with \(\frac{d^2y}{dx^2}\): $$x^{2}\frac{d^2y}{dx^2} + 7x\frac{dy}{dx} + 5y = 0$$ Now we can make the substitution \(y(x) = x^r\) and differentiate twice to obtain: First derivative: \(\frac{dy}{dx}=rx^{r-1}\) Second derivative: \(\frac{d^2y}{dx^2}=r(r-1)x^{r-2}\) Plug these into the homogeneous equation: $$x^{2}[r(r-1)x^{r-2}] + 7x[rx^{r-1}] + 5[x^r] = 0$$ Simplify by dividing both sides by \(x^{r-2}\): $$r(r-1)x^{2} + 7r x^3 + 5x^5 = 0$$ Now, we look for the values of \(r\) for which this equation is true.

Solving for r and finding the complementary function

The equation simplifies to: $$r(r-1) + 7r + 5=0$$ Now we can solve for \(r\) in terms of a quadratic equation: $$r^2 - r + 7r + 5 = 0$$ $$r^2 + 6r + 5 = 0$$ Factor the quadratic equation: $$(r+1)(r+5) =0$$ So we have two possible values for \(r\), \(r_1 = -1\) and \(r_2=-5\). The complementary function is, therefore, a linear combination of these two solutions: $$y_c(x) = C_1x^{-1} + C_2x^{-5}$$

Finding the particular solution

Now, we need to find a particular solution of the given non-homogeneous equation: $$x^{2}y'' + 7xy' + 5y = x$$ We guess a solution of the form \(y_p(x) = Ax^2 + Bx\). Differentiating twice, we obtain: First derivative: \(y_p'(x) = 2Ax + B\) Second derivative: \(y_p''(x) = 2A\) Plug these into the non-homogeneous equation: $$x^2(2A) + 7x(2Ax + B) + 5(Ax^2+ Bx) = x$$ Simplify: $$2Ax^4 + (14A+5B)x^3 + 10Ax^2 + 5B = x$$ Comparing coefficients of the powers of \(x\), we obtain the following system of equations: $$2A = 0 \Rightarrow A = 0$$ $$14A+5B=0 \Rightarrow 5B = 0 \Rightarrow B = \frac{1}{5}$$ Thus, the particular solution is: $$y_p(x) = \frac{1}{5}x$$

Finding the general solution

Now that we have the complementary function and particular solution, we can find the general solution by adding these two: $$y(x) = y_c(x) + y_p(x) = C_1x^{-1} + C_2x^{-5} + \frac{1}{5}x$$ The general solution of the given equation is: $$y(x) = C_1x^{-1} + C_2x^{-5} + \frac{1}{5}x$$

Key concepts

Complementary Function

When solving second order differential equations, the complementary function (CF), also known as the homogeneous solution, is critical to understanding the behavior of the system without external forces or inputs. In our example, the CF is found by solving the homogeneous equation associated with the original differential equation—meaning, we set the right-hand side to zero.

The approach used here involves substituting a solution of the form \(y(x) = x^r\), known as Euler's method for differential equations, into the homogeneous equation and determining the values of \(r\) that satisfy it. We then obtained two separate solutions corresponding to each value of \(r\), resulting in \(y_c(x) = C_1x^{-1} + C_2x^{-5}\). Here, \(C_1\) and \(C_2\) are constants to be determined by initial conditions or boundary values.

Understanding the complementary function is foundational in solving differential equations as it represents the natural response of the system without external input.

Particular Solution

The particular solution (PS) to a differential equation accounts for the effects of the non-homogeneous part (external input) of the equation. In contrast to the complementary function, which considers the system's behavior in the absence of external forces, the particular solution reflects how the system responds specifically to the external input—in this exercise, \(x\).

To find the particular solution, we make a reasonable guess based on the form of the non-homogeneous term, which was \(y_p(x) = Ax^2 + Bx\) in our case. After substituting and comparing coefficients, we deduced that \(A=0\) and \(B=\frac{1}{5}\), leading to the specific solution \(y_p(x) = \frac{1}{5}x\). It's critical for students to grasp that the form of the guess must be informed by the type of non-homogeneous term presented in the differential equation.

Second Order Differential Equations

Second order differential equations are equations that involve the second derivative of a function. They are prominent in physics and engineering as they often model systems with acceleration or curvature, such as vibrations and oscillations.

The general form of a second order differential equation is \(a(x)y'' + b(x)y' + c(x)y = g(x)\), where \(y''\) denotes the second derivative of \(y\) with respect to \(x\), and \(a(x)\), \(b(x)\), and \(c(x)\) are functions of \(x\) that can vary or be constant coefficients, while \(g(x)\) represents an external force or input.

Our specific exercise presented a situation in which \(a(x)\), \(b(x)\), and \(c(x)\) all had specific relationships to \(x\), and we were tasked to solve \(x^2y'' + 7xy' + 5y = x\), showcasing a vital application for these types of equations.

Characteristic Equation

The characteristic equation arises when using the method of solving linear differential equations with constant coefficients. It's an essential component because it establishes a polynomial whose roots directly determine the complementary function of the differential equation.

For our exercise, after employing a substitution for \(y(x)\), we obtained a quadratic equation in \(r\) which is a characteristic equation of our homogeneous differential equation. We solved \(r^2 + 6r + 5 = 0\) to find the characteristic roots \(r_1 = -1\) and \(r_2 = -5\). These roots are invaluable because they tell us the precise form of the complementary function that satisfies the homogeneous part of our original differential equation.

General Solution

The general solution of a differential equation is the sum of its complementary function and particular solution. It represents the most comprehensive expression for the behavior of the system under study, covering both the natural response and the forced response.

In our example, combining the complementary function \(y_c(x) = C_1x^{-1} + C_2x^{-5}\) and the particular solution \(y_p(x) = \frac{1}{5}x\) gives us the general solution: \[y(x) = C_1x^{-1} + C_2x^{-5} + \frac{1}{5}x\]. The constants \(C_1\) and \(C_2\) allow the solution to be tailored to fit specific initial conditions, highlighting the importance of these constants in crafting a solution that reflects real-world constraints.