Question

Find the general solution, given that \(y_{1}\) satisfies the complementary equation. As a byproduct, find a fundamental set of solutions of the complementary equation. \(x^{2} y^{\prime \prime}-3 x y^{\prime}+4 y=4 x^{4} ; \quad y_{1}=x^{2}\)

Short answer

Question: Find the general solution to the differential equation \(x^2 y'' - 3xy' + 4y = 4x^4\), given that \(y_1 = x^2\) is a solution to the homogeneous version of this equation. Answer: The general solution to the given differential equation is \(y(x) = C_1 x^2 + C_2 x^4 - \frac{1}{2}x^6 + 2x^5\), where \(C_1\) and \(C_2\) are arbitrary constants.

Step-by-step solution

Find the complementary function of the homogeneous version of the given equation

The homogeneous form of the given equation is: $$x^2 y'' - 3xy' + 4y = 0$$ We already have one solution, \(y_1 = x^2\). To find the second solution, \(y_2\), we'll use the method of reduction of order. Assume \(y_2 = v \cdot y_1 = vx^2\), where \(v\) is a function of \(x\). First, compute the derivatives of \(y_2\): $$y_2' = v'x^2 + 2vx \quad \text{and} \quad y_2'' = v''x^2 + 4v'x + 2v$$ Substitute these derivatives into the homogeneous equation: $$x^2(v''x^2 + 4v'x + 2v) - 3x(v'x^2 + 2vx) + 4(vx^2) = 0$$ Simplify the equation: $$x^2v''x^2 + 4x^2v'x + 2x^2v - 3x^3v' - 6x^2v + 4vx^2 = 0$$ Divide by \(x^2\): $$x^2v'' + 4xv' - 3x^2v' + 2v - 6v + 4v = 0$$ Now, we have a first-order linear equation in terms of \(v'\): \((x^2 - 3x^2)v' + (4x)(v - 2v + v) = 0\) Simplify: $$-2x^2v' + 4xv = 0$$ Divide by \(-2x\) to obtain the separable equation: $$v' = 2 \cdot \frac{v}{x}$$ Now, we can integrate to find \(v\): $$\int \frac{dv}{v} = \int 2 \cdot \frac{dx}{x}$$ Take the integral of both sides: $$\ln|v| = 2 \ln|x| + C_1$$ Solve for \(v\): $$v = C_2 x^2$$ Now we can find the solution, \(y_2\): $$y_2 = vx^2 = C_2 x^4$$ Thus, we have a fundamental set of solutions for the homogeneous equation: \(\{x^2, x^4\}\).

Find the particular solution using the variation of parameters

To find the particular solution, we'll use the method of variation of parameters. Assume that the particular solution is in the form: $$Y_p = u_1(x)x^2 + u_2(x)x^4$$ We need to find the functions \(u_1(x)\) and \(u_2(x)\). Compute the Wronskian of the complementary function: $$W = \begin{vmatrix} x^2 & x^4 \\ 2x & 4x^3 \end{vmatrix} = 4x^5 - 2x^5 = 2x^5$$ The Wronskian is nonzero, so we can proceed to find \(u_1(x)\) and \(u_2(x)\): $$u_1'(x) = \frac{-y_2 \cdot g(x)}{W} = \frac{-x^4(4x^4)}{2x^5} = -2x^3$$ $$u_2'(x) = \frac{y_1 \cdot g(x)}{W} = \frac{x^2(4x^4)}{2x^5} = 2$$ Now, we can integrate to find \(u_1(x)\) and \(u_2(x)\): $$u_1(x) = \int -2x^3 dx = -\frac{1}{2}x^4 + C_3$$ $$u_2(x) = \int 2 dx = 2x + C_4$$ Now, we can write the particular solution \(Y_p\): $$Y_p = (-\frac{1}{2}x^4 + C_3)x^2 + (2x + C_4)x^4$$ Since we only need one particular solution, we can set the constants, \(C_3\) and \(C_4\), to zero: $$Y_p = -\frac{1}{2}x^6 + 2x^5$$

Find the general solution

The general solution is the sum of the complementary function and the particular solution: $$y(x) = C_1 x^2 + C_2 x^4 - \frac{1}{2}x^6 + 2x^5$$ Thus, the general solution to the given nonhomogeneous equation is: $$y(x) = C_1 x^2 + C_2 x^4 - \frac{1}{2}x^6 + 2x^5$$

Key concepts

Complementary Function

In the context of differential equations, a complementary function represents the solution to the associated homogeneous differential equation, which is derived by setting the non-homogeneous part of the equation to zero. To solve the homogeneous equation, we aim to find the complementary function that satisfies the equation.
Given the equation:\[x^2 y'' - 3xy' + 4y = 0\]We know that one known solution is provided as \(y_1 = x^2\). The task is to identify the second solution, \(y_2\), using the **Reduction of Order** method explained below. Once identified, the general solution of the homogeneous equation, or complementary function, will be in the form:

• \(C_1y_1 + C_2y_2 = C_1 x^2 + C_2 x^4\)

This solution, expressed through arbitrary constants \(C_1\) and \(C_2\), encompasses all possible solutions to the homogeneous part of the differential equation.

Reduction of Order

**Reduction of Order** is a technique used to find a second linearly independent solution to a homogeneous differential equation when one solution is already known. In this method, we assume a solution in the form \(y_2 = v(x) imes y_1(x)\), where \(y_1\) is the known solution, and \(v(x)\) is an unknown function to determine.
For our problem, given \(y_1 = x^2\), we assume:

• \(y_2 = v(x) imes x^2\)
• Calculate derivatives: \(y_2' = v'x^2 + 2vx\) and \(y_2'' = v''x^2 + 4v'x + 2v\)

Substitute these into the homogeneous equation:\[x^2(v''x^2 + 4v'x + 2v) - 3x(v'x^2 + 2vx) + 4(vx^2) = 0\]This simplifies into a separable first-order differential equation in terms of \(v'\). Solving this gives:\[v = C_2 x^2\]Thus, the second solution is \(y_2 = C_2 x^4\), completing the fundamental set \(\{x^2, x^4\}\).

Variation of Parameters

The **Variation of Parameters** method is employed to find a particular solution for non-homogeneous differential equations. It leverages the solutions of the corresponding homogeneous equation. For our problem, we express the particular solution in the form:
\[Y_p = u_1(x)x^2 + u_2(x)x^4\]
Here, \(u_1(x)\) and \(u_2(x)\) are functions that need to be determined. The Wronskian, \(W\), helps in this calculation:

• Compute the Wronskian of \(y_1 = x^2\) and \(y_2 = x^4\):
  \[W = \begin{vmatrix} x^2 & x^4 \ 2x & 4x^3 \end{vmatrix} = 2x^5\]
• Find the derivatives of \(u_1\) and \(u_2\):
  • \(u_1' = \frac{-y_2 imes g(x)}{W} = -2x^3\)
  • \(u_2' = \frac{y_1 imes g(x)}{W} = 2\)

Integrate these expressions to find \(u_1(x)\) and \(u_2(x)\), leading to:

• \(u_1 = -\frac{1}{2}x^4 + C_3\)
• \(u_2 = 2x + C_4\)

With constants typically set to zero for simplicity, the particular solution \(Y_p\) becomes:\[Y_p = -\frac{1}{2}x^6 + 2x^5\]

Homogeneous Equation

A **Homogeneous Equation** in the context of differential equations is one where there is no term independent of the function or its derivatives. Simply put, every term has the dependent variable or its derivatives involved. The original exercise involves converting a non-homogeneous equation into a homogeneous form to find the complementary solution.
For example, if you start with:
\[x^2 y'' - 3xy' + 4y = 4x^4\]
The homogeneous counterpart is derived by removing the non-homogeneous part \(4x^4\):
\[x^2 y'' - 3xy' + 4y = 0\]Solving the homogeneous equation is crucial because it serves as the basis for constructing the general solution of the non-homogeneous equation. The complementary function derived from this homogeneous equation captures the behavior of the solutions based entirely on their internal structure without any external forcing functions.