Question

Find the general solution. $$ y^{\prime \prime \prime}+2 y^{\prime \prime}+y^{\prime}=-2 e^{-x}\left(7-18 x+6 x^{2}\right) $$

Short answer

Question: Find the general solution to the following third-order linear ordinary differential equation: $$ y''' + 2y'' + y' = -2 e^{-x}\left(7 - 18x + 6x^2\right) $$ Answer: The general solution to the given third-order linear ordinary differential equation is: $$ y = (C_1 + C_2 e^{-x} + C_3 x e^{-x}) + (x^2 - 2x + 4)e^{-x} $$ where \(C_1\), \(C_2\), and \(C_3\) are arbitrary constants.

Step-by-step solution

Solve the homogeneous equation

The given ODE can be written as: $$ D^3y + 2D^2y + Dy = 0 $$ Where \(D = \frac{d}{dx}\). The homogeneous equation is: $$ D^3y + 2D^2y + Dy = 0 $$ The characteristic equation is: $$ r^3 + 2r^2 + r = 0 $$ Factor out the common term, \(r\): $$ r(r^2 + 2r + 1) = 0 $$ This factors further into: $$ r(r+1)^2 = 0 $$ The roots are \(r=0\), with multiplicity 1, and \(r=-1\), with multiplicity 2.

Find the complementary function

Based on the roots of the characteristic equation, the complementary function is: $$ y_c = C_1 e^{0x} + C_2 e^{-x} + C_3 xe^{-x} $$ Which simplifies to: $$ y_c = C_1 + C_2 e^{-x} + C_3 xe^{-x} $$

Determine the form of the particular solution

The non-homogeneous term of the given ODE is: $$ -2 e^{-x}\left(7 - 18x + 6x^2\right) $$ This suggests that the form of the particular solution should be: $$ y_p = (Ax^2 + Bx + C)e^{-x} $$

Differentiate the particular solution

Compute the first, second, and third derivatives of the particular solution: First derivative: $$ y_p' = e^{-x}(-Ax^2 - (2A+B)x + (A-2B-C)) $$ Second derivative: $$ y_p'' = e^{-x}(2Ax^2 + 3(2A+B)x - (2A+4B+3C)) $$ Third derivative: $$ y_p''' = e^{-x}(-6Ax^2 - 6(2A+B)x + (-6A+12B+3C)) $$

Substitute the particular solution derivatives into the given ODE

Substitute \(y_p'''\), \(y_p''\), and \(y_p'\) into the given ODE to find the coefficients \(A\), \(B\), and \(C\): $$ (-6Ax^2 - 6(2A+B)x + (-6A+12B+3C)) + 2(2Ax^2 + 3(2A+B)x - (2A+4B+3C)) + (-Ax^2 - (2A+B)x + (A-2B-C)) =\\ -2 e^{-x}\left(7 - 18x + 6x^2\right) $$ Comparing the coefficients of corresponding terms, we get: $$ 6A = -6, \quad -12A - 3B = 18, \quad 6A - 12B - 3C = -14 $$

Calculate the coefficients

Solve the system of linear equations for the coefficients \(A\), \(B\), and \(C\): $$ A = 1, \quad B = -2, \quad C = 4 $$

Write the particular solution

Substitute the found coefficients back into the form of the particular solution: $$ y_p = (x^2 - 2x + 4)e^{-x} $$

Combine the complementary function and the particular solution

The general solution of the given ODE is the sum of the complementary function and the particular solution: $$ y = y_c + y_p = (C_1 + C_2 e^{-x} + C_3 x e^{-x}) + (x^2 - 2x + 4)e^{-x} $$

Key concepts

General Solution of ODE

When approaching ordinary differential equations (ODEs), the ultimate goal is to find the general solution, which represents all possible solutions to the equation. A general solution contains arbitrary constants, which correspond to the number of times you have to differentiate the function to get the original function back - essentially the order of the ODE.

For the given ODE, which is third-order, the general solution will include three arbitrary constants. This because the higher the order of the derivative, the more initial conditions you can satisfy. Thus, a third-order equation allows for three initial conditions. The general solution is a combination of two parts: the complementary function (y_c), which solves the associated homogeneous equation, and the particular solution (y_p), which solves the non-homogeneous equation. These two parts are summed to give the most comprehensive solution possible, considering all scenarios covered by the differential equation.

Characteristic Equation

Solving higher-order linear homogeneous ordinary differential equations involves a critical step: forming and solving the characteristic equation. This auxiliary equation arises from replacing each derivative of the dependent variable with a term involving a hypothetical exponent r.

For an equation such as the third-order one given, we substitute each derivative by powers of r, for example, D³y becomes r³, which leads to the algebraic characteristic equation. This equation helps us to determine the nature of the solution (whether it involves real numbers, complex numbers, or repeated roots) and thus significantly simplify the finding of the complementary function. In our exercise, solving the characteristic equation gives us the roots, which are the key to constructing the complementary function.

Complementary Function

The complementary function, often denoted as y_c, specifically addresses the homogeneous portion of an ODE, which is the part without the non-homogeneous term (any terms not multiplied by the function or its derivatives). The complementary function is built from the roots of the characteristic equation.

In the given problem, the characteristic equation yielded the roots r = 0, and r = -1 with multiplicity 2. The complementary function is thus a linear combination of the exponential functions e^0x (which simply is 1), e^-x, and xe^-x, each multiplied by an arbitrary constant – C₁, C₂, and C₃ respectively. These constants will later be determined by initial conditions or boundary values provided.

Particular Solution

While the complementary function relates to the homogeneous equation, the particular solution, y_p, is the specific solution that accounts for the non-homogeneous part of the ODE. It's 'particular' because it corresponds to one particular set of conditions outlined in the non-homogeneous part of the equation.

In our example, the ODE includes a right-hand side that is non-homogeneous, suggesting a specific form for y_p that resembles the non-homogeneous part. The form should be guessed in such a way to avoid overlap with the complementary function. After determining this form, it's necessary to differentiate it, substitute back into the ODE, and solve for any unknown constants. Finally, these constants are plugged back into the presumed form to attain the particular solution. The particular solution, along with the complementary function, comprises the final, general solution.