Question

For each differential equation, (a) Find the complementary solution. (b) Formulate the appropriate form for the particular solution suggested by the method of undetermined coefficients. You need not evaluate the undetermined coefficients. $$ y^{\prime \prime \prime}-4 y^{\prime \prime}+4 y^{\prime}=t^{3}+4 t^{2} e^{2 t} $$

Short answer

#Question# Find the complementary solution of the following differential equation and then formulate the appropriate form for the particular solution using the method of undetermined coefficients without actually evaluating the undetermined coefficients: $$y^{\prime \prime \prime}-4y^{\prime \prime}+4y^{\prime}=t^{3}+4t^{2}e^{2t}$$ #Answer# The complementary solution is: $$y_c(x)=C_{1}+C_{2}e^{2x}+C_{3}xe^{2x}$$ The appropriate form of the particular solution is: $$Y_p(t)=t^3(at^3+bt^2+ct+d)+4t^2e^{2t}(At^2+Bt+C)e^{2t}$$

Step-by-step solution

Identify the homogeneous equation and its characteristic equation

From the given differential equation, the homogeneous equation is: $$y^{\prime \prime \prime}-4y^{\prime \prime}+4y^{\prime}=0$$ The characteristic equation is obtained by substituting \(y=e^{mx}\) in the homogeneous equation, which yields the polynomial: $$m^{3}-4m^{2}+4m=0$$

Solve the characteristic equation to find characteristic roots

Factor the characteristic equation: $$m(m^{2}-4m+4)=0$$ $$(m)(m-2)^{2}=0$$ Roots are: \(m=0\), \(m=2\), and \(m=2\) (repeated root).

Form the complementary solution using the characteristic roots

The complementary solution is formed by using the characteristic roots and the general equation for a third-order linear homogeneous differential equation with repeated roots. For this case, it is: $$y_c(x)=C_{1}e^{0x}+C_{2}e^{2x}+C_{3}xe^{2x}$$ Taking \(e^{0x}=1\), the complementary solution becomes: $$y_c(x)=C_{1}+C_{2}e^{2x}+C_{3}xe^{2x}$$ This is the complementary solution required for part (a).

Identify the non-homogeneous terms and adjust for any overlap with the complementary solution

In the given differential equation, the non-homogeneous terms are \(t^3\) and \(4t^2e^{2t}\). There are no overlaps with the complementary solution, so we can proceed without any modifications.

Formulate the appropriate form for the particular solution using the method of undetermined coefficients

For the polynomial part, \(t^3\), the appropriate form is: $$Y_p=t^3(at^3+bt^2+ct+d)$$ For the exponential part, \(4t^2e^{2t}\), the appropriate form is: $$Y_p=4t^2e^{2t}(At^2+Bt+C)e^{2t}$$ Combining both parts, we have the appropriate form of the particular solution: $$Y_p(t)=t^3(at^3+bt^2+ct+d)+4t^2e^{2t}(At^2+Bt+C)e^{2t}$$ This is the appropriate form of the particular solution for part (b).

Key concepts

Complementary Solution

In solving differential equations, the complementary solution is essential in understanding the behaviors of the system described by the equation. The complementary solution, denoted as \( y_c(x) \), applies specifically to the homogeneous part of a differential equation. It represents the general solution derived from the associated homogeneous equation.

The homogeneous equation ignores the non-homogeneous terms (i.e., terms on the right side of the equal sign) and looks like \( y^{\prime \prime \prime} - 4y^{\prime \prime} + 4y^{\prime} = 0 \). For this kind of equation, we seek solutions of the form \( e^{mx} \). By solving the characteristic equation, which is derived from the homogeneous equation, we can find suitable values of \( m \) to use in the solution. In this case, the complementary solution becomes:
\[y_c(x) = C_1 + C_2 e^{2x} + C_3 x e^{2x}\]
Here, \( C_1, C_2, \) and \( C_3 \) are constants determined by boundary or initial conditions. This solution is the building block for the general solution of the entire differential equation.

Characteristic Equation

The characteristic equation stems from transforming the homogeneous differential equation into a polynomial equation. To do this, we express the solution in the form \( y = e^{mx} \) and then substitute this expression into the differential equation. This leads us to the characteristic equation, which in our example is:
\[m^3 - 4m^2 + 4m = 0\]
Solving this equation involves finding the roots, \( m \), which are critical in determining the form of the complementary solution. In this particular exercise, factoring the characteristic equation yields the roots:

• \( m = 0 \)
• \( m = 2 \) (repeated root)

These roots dictate the terms in our complementary solution. If the roots are repeated, the form of the solution changes slightly to accommodate those repetitions, as seen in terms like \( x e^{2x} \). Understanding and solving the characteristic equation is a cornerstone of handling linear differential equations.

Undetermined Coefficients

The method of undetermined coefficients is a powerful technique used to find the particular solution of a non-homogeneous differential equation. This approach involves guessing a form for the particular solution based on the type of non-homogeneous term and coefficients.

The key to this method is selecting the correct form to ensure that every component of the non-homogeneous part is addressed. For instance, given the non-homogeneous terms like \( t^3 \) and \( 4t^2e^{2t} \), we propose specific forms for the particular solution:

• For \( t^3 \), we use \( Y_p = a t^3 + b t^2 + ct + d \).
• For \( 4t^2 e^{2t} \), we choose \( Y_p = (At^2 + Bt + C)e^{2t} \).

Combining these, our assumed particular solution can be written as an entity that accommodates all variations in the non-homogeneous terms.
After selecting the form, these coefficients ('a', 'b', 'c', 'd', 'A', 'B', 'C') are determined by substituting the assumed solution into the differential equation and solving for each coefficient to balance both sides of the equation.

Homogeneous Differential Equation

A homogeneous differential equation is a type where every term is a function of the unknown variable and its derivatives, set to zero. This contrasts with a non-homogeneous equation that has additional standalone terms. In solving such equations, the homogeneous part takes the spotlight initially, assisting in forming part of the overall solution.

The equation \( y^{\prime \prime \prime} - 4y^{\prime \prime} + 4y^{\prime} = 0 \) is a classic example of a homogeneous differential equation. Its solutions involve exponential functions derived from the characteristic equation's roots.

Understanding these equations is imperative because they allow us to solve for the complementary solution.

• These solutions showcase general behaviors common to all solutions.
• They determine stability and oscillatory properties of the system.

The practice of working with homogeneous equations forms the backbone of more complex problem-solving in differential equations, providing essential insight into the system being modeled.