Question

(a) Find the general solution \(y_{\mathrm{h}}\) of the homogeneous differential equation \(\frac{d y}{d x}+2 x y=0\) (b) Show that the general solution of the nonhomogeneous equation \(\frac{d y}{d x}+2 x y=3 e^{-x^{2}}\) is equal to the solution \(y_{b}\) in part (a) plus a particular solution to the nonhomogeneous equation.

Short answer

#Answer# The general solution of the given nonhomogeneous differential equation is: \(y(x) = Ce^{-x^2} - 3xe^{-x^2}\)

Step-by-step solution

Identify the coefficients of the homogeneous differential equation

The given homogeneous differential equation is: \(\frac{dy}{dx} + 2xy = 0\) In this equation, the coefficients are: P(x) = 2x

Find the integrating factor

Now, let's find the integrating factor by integrating the coefficient P(x): Integrating factor (IF) = \(e^{\int P(x) dx}= e^{\int 2x dx}\) IF = \(e^{x^2}\)

Find the general solution

The general solution \(y_h\) of the homogeneous differential equation is found by multiplying y by the integrating factor and integrating: \( y_h(x) e^{x^2} = \int 0 \cdot e^{x^2} dx\) \( y_h(x) e^{x^2} = C \), where C is the constant of integration So, the general solution \(y_h\) of the homogeneous differential equation is: \( y_h(x) = Ce^{-x^2}\) ##Part (b)##

Use an educated guess to find a particular solution

Now, we need to find a particular solution of the nonhomogeneous differential equation: \(\frac{dy}{dx} + 2xy = 3e^{-x^2}\) We can make an educated guess that the particular solution \(y_p\) would be in the form: \( y_p(x) = A(x) e^{-x^2}\), where A(x) is an unknown function To find A(x), differentiate \(y_p\) and substitute it into the nonhomogeneous equation.

Substitute and solve for A(x)

Substitute the educated guess and its derivative into the equation and solve for A(x). \(\frac{d(A(x) e^{-x^2})}{dx}+2x(A(x) e^{-x^2})=3 e^{-x^2}\) After simplifying and cancelling out \(e^{-x^2}\), we get: \(\frac{dA}{dx}(x) - 2xA(x) = 3\) Now, take IF = \(e^{\int (-2x) dx} = e^{-x^2}\) and multiply both sides: \(A'(x)e^{-x^2} - 2xA(x)e^{-x^2} = 3e^{-x^2}\) But we know that \(A'(x)e^{-x^2} - 2xA(x)e^{-x^2} = \frac{d(A(x)e^{-x^2})}{dx}\) So, \(\frac{d(A(x)e^{-x^2})}{dx} = 3e^{-x^2}\) Now, integrate both sides and get: \(A(x)e^{-x^2} = -3xe^{-x^2} + C\) So, \(A(x) = -3x + Ce^{x^2}\) But, we only need the particular solution, and we already have the homogeneous part from part (a): \(y_p(x) = A(x)e^{-x^2} = -3xe^{-x^2}+ Ce^{x^2}e^{-x^2}\) Since we need only the particular solution, we can discard the \(Ce^{x^2}e^{-x^2}\) term: \(y_p(x) = -3xe^{-x^2}\)

Combine the solutions

Both the homogeneous solution and the particular solution are found. Now we can combine them to find the general solution for the nonhomogeneous differential equation: \(y(x) = y_h(x) + y_p(x)\) \(y(x) = Ce^{-x^2} - 3xe^{-x^2}\) So, the general solution of the given nonhomogeneous differential equation is: \(y(x) = Ce^{-x^2} - 3xe^{-x^2}\)

Key concepts

Homogeneous Differential Equations

Homogeneous differential equations are foundational in the study of ordinary differential equations. In essence, a homogeneous differential equation is characterized by an expression set equal to zero. The general format for a first-order homogeneous differential equation is \( \frac{dy}{dx} + P(x)y = 0 \), where \( P(x) \) represents a function of \( x \) alone.

The solution to such an equation lies in finding an integrating factor that will allow us to get the solution in a simpler form. For the given problem \( \frac{dy}{dx} + 2xy = 0 \), the integrating factor is found to be \( e^{x^2} \), which stems from the coefficient \( 2x \). The essence of this factor is to transform the differential equation into an exact one, making integration straightforward. As a result, the general solution \( y_h(x) = Ce^{-x^2} \) is attained, where \( C \) is the constant of integration. This expression represents an infinite family of curves, each corresponding to a different value for \( C \), which can be determined by an initial condition if provided.

Nonhomogeneous Differential Equations

In contrast to their homogeneous counterparts, nonhomogeneous differential equations include a nonzero function on one side of the equation. Generally written as \( \frac{dy}{dx} + P(x)y = Q(x) \) where \( Q(x) \) is the nonhomogeneous or forcing term. These equations are often more complex to solve since the solution involves finding both a homogeneous solution and a particular solution that accounts for the \( Q(x) \) term.

The given nonhomogeneous equation \( \frac{dy}{dx} + 2xy = 3e^{-x^2} \) requires us to first understand the homogeneous solution—already found as \( y_h(x) = Ce^{-x^2} \)—and then find a particular solution \( y_p(x) \) for the nonhomogeneous part. The overall general solution for the nonhomogeneous equation is the sum of these two parts, demonstrating that the nature of the solutions to these equations is additive.

Integrating Factor

An integrating factor is a clever device used in the solving of linear first-order differential equations. It’s essentially a function, which, when multiplied by the original equation, renders it integrable in a more straightforward manner.

The integrating factor for a first-order equation \( \frac{dy}{dx} + P(x)y = Q(x) \) is commonly denoted as \( e^{\int P(x) dx} \), mirroring the effect of having integrated the \( P(x) \) term directly. After finding the integrating factor, the equation is multiplied through by this factor, leading to a left-hand side that can be expressed as the derivative of the product of the integrating factor and \( y(x) \), simplifying the path to the solution. In the exercise, we used this method to transform both the homogeneous and nonhomogeneous equations into a form that was easy to integrate.

Particular Solution

When tackling nonhomogeneous differential equations, the particular solution \( y_p \) is the part that specifically addresses the non-zero term in the equation. Unlike the general solution of the homogeneous equation, which includes a constant of integration \( C \), the particular solution is a unique function that does not include any arbitrary constants—its shape is solely determined by the function \( Q(x) \) in the nonhomogeneous term.

To find \( y_p \), we frequently use methods of undetermined coefficients or variation of parameters, depending on the form of \( Q(x) \). An educated guess, also known as an 'Ansatz', can sometimes streamline finding the particular solution, as seen in our problem where the particular solution is constructed as \( y_p(x) = A(x)e^{-x^2} \). This strategy requires us to then determine the function \( A(x) \) by differentiating the guess, substituting back into the original equation, and solving for the unknowns. It's pivotal to comprehend that the particular solution greatly matters since it ensures that the complete general solution accounts for the nonhomogeneous part of the differential equation.