Question

Use variation of parameters to find a particular solution. $$ y^{\prime \prime}-2 y^{\prime}+y=14 x^{3 / 2} e^{x} $$

Short answer

The method used to find a particular solution to a nonhomogeneous differential equation is called the variation of parameters method. This method involves introducing suitable functions to multiply with the homogeneous solutions in order to satisfy the original equation. The most important part is determining the parameters to find the particular solution, which is done through a series of steps involving integration and imposing conditions on the functions.

Step-by-step solution

Identify the differential equation and its homogeneous form

The given nonhomogeneous differential equation is $$ y^{\prime \prime}-2 y^{\prime}+y=14 x^{3 / 2} e^{x}. $$ The corresponding homogeneous differential equation is $$ y^{\prime \prime}-2 y^{\prime}+y=0. $$

Find the complementary function (general solution to the homogeneous equation)

The homogeneous equation has a characteristic equation given by $$ r^2 - 2r + 1 = 0. $$ This can be factored as $$ (r - 1)^2 = 0. $$ As there is only one distinct root, \(r = 1\), the general solution of the homogeneous equation is given by $$ y_c(x) = C_1 e^x + C_2 x e^x. $$

Apply the variation of parameters method

To find a particular solution using variation of parameters, we proceed to introduce two arbitrary functions u(x) and v(x) such that $$ y_p(x) = u(x) e^x + v(x) x e^x. $$ Now, we compute the first and second derivatives of \(y_p(x)\): $$ y_p'(x) = u'(x) e^x + v'(x) x e^x + v(x) e^x, $$ and $$ y_p''(x) = u''(x) e^x + v''(x) x e^x + 2 v'(x) e^x. $$

Plug the derivatives into the nonhomogeneous equation

Replace \(y_p\), \(y_p'\), and \(y_p''\) in the non-homogeneous equation with their expressions: $$ [u''(x) e^x + v''(x) x e^x + 2 v'(x) e^x] - 2[u'(x) e^x + v'(x) x e^x + v(x) e^x] + [u(x) e^x + v(x) x e^x] = 14 x^{3/2} e^x. $$

Simplify and find conditions on u and v

By grouping terms with the same functions, we get $$ [u''(x) - 2u'(x) + u(x)]e^x + [v''(x)x - 2v'(x)x + 2v'(x) - 2v(x) + vx]e^x = 14x^{3/2} e^x. $$ Since \(y_c(x) = C_1 e^x + C_2 x e^x\) is already the general solution of the homogeneous equation, we shouldn't have terms containing \(u\) and \(v\) without their derivatives. So we impose conditions: $$ u''(x) - 2u'(x) + u(x) = 0, $$ and $$ v''(x)x - 2v'(x)x + 2v'(x) - 2v(x) + vx = 14x^{3/2}. $$ The condition on \(u\) is already satisfied by the choice of \(u(x)e^x\). Now we need to find \(v(x)\) satisfying the second condition.

Integrate to find v(x)

The condition on \(v\) is a first-order differential equation for \(v'(x)\), since we can rewrite it as: $$ v''(x)x - 2v'(x)(x-1) - 2v(x) + vx = 14x^{3/2}. $$ We now integrate twice to find \(v(x)\): 1. Integration $$ \int (v''(x)x - 2v'(x)(x-1)) dx = \int (14x^{3/2} + 2v(x) - vx) dx. $$ Then, $$ v'(x)(x-1) - v'(x) = 14\frac{x^{5/2}}{5/2} + 2v(x) + \frac{v(x)x^2}{2} + C_1. $$ 2. Integration $$ \int (v'(x)(x-1) - v'(x)) dx = \int (14\frac{x^{5/2}}{5/2} + 2v(x) + \frac{v(x)x^2}{2} + C_1) dx. $$ This gives $$ v(x) = \frac{14x^{7/2}}{21} + C_1 \ln x + C_2. $$

Express the particular solution

We have found the functions \(u(x) = e^x\) and \(v(x) = \frac{14x^{7/2}}{21} + C_1 \ln x + C_2\). Recall that $$ y_p(x) = u(x) e^x + v(x) x e^x. $$ Substituting \(u(x)\) and \(v(x)\), we get $$ y_p(x) = e^x + x e^x \left(\frac{14x^{7/2}}{21} + C_1 \ln x + C_2\right). $$ The particular solution is the one without constants \(C_1\) and \(C_2\): $$ y_p(x) = e^x + x e^x \frac{14x^{7/2}}{21}. $$

Key concepts

Differential Equations

Differential equations are mathematical equations that describe the relationship between a function and its derivatives. They play a crucial role in various fields such as physics, engineering, and economics because they model the behavior of dynamic systems.

When solving differential equations, one of your main goals is to find a function (or a set of functions) that satisfies the given equation. This process usually involves identifying the type of differential equation you're dealing with, such as whether it's ordinary or partial, linear or nonlinear, and then applying the appropriate method to find a solution.

In our exercise, we encounter an ordinary differential equation because it involves derivatives with respect to only one variable, which in this case is 'x'. The equation includes second-order derivatives, denoted by the notation \( y'' \), and it is linear since the terms involving \( y \), \( y' \), and \( y'' \) do not multiply or divide each other and no functions of these variables are involved.

Particular Solution

The particular solution to a differential equation is a solution that fits both the form of the differential equation and the specific initial conditions or boundary conditions given in the problem. Essentially, it’s the 'special' solution that satisfies all the conditions imposed.

To find a particular solution, often we need to first find the general solution, which includes arbitrary constants. From the general solution, we apply methods such as 'variation of parameters' that allow us to consider the specific nonhomogeneous part of the equation (in this case, \(14 x^{3 / 2} e^{x}\)).

The 'variation of parameters' method relies on finding functions which, when substituted into the equation, transform it into an identity. The final form of a particular solution doesn’t include the arbitrary constants (\(C_1\), \(C_2\), etc.), as it relates explicitly to the nonhomogeneous part of the original equation.

Homogeneous Equations

Homogeneous differential equations are those in which every term is a function of the dependent variable (here, 'y') and its derivatives. These equations are simpler to solve compared to their nonhomogeneous counterparts because they possess the property that if \(y(x)\) is a solution, so is \(c \times y(x)\) for any constant 'c'.

In the context of our exercise, by removing the nonhomogeneous part (\(14 x^{3 / 2} e^{x}\)), we obtain a homogeneous equation \( y'' - 2y' + y = 0 \), which we refer to as the complementary or homogeneous equation. Solving it gives us the general solution or complementary function \(y_c(x)\), which contains arbitrary constants and represents the general behavior of the system without external forces.

The crucial aspect of solving a nonhomogeneous equation is to combine the homogeneous solution with a particular solution to compose the most general form of the solution, encompassing both the inherent system dynamics and the external forces.