Question

Exercises \(31-36\) treat the equations considered in Examples \(5.4 .1-5.4 .6 .\) Substitute the suggested form of \(y_{p}\) into the equation and equate the resulting coefficients of like functions on the two sides of the resulting equation to derive a set of simultaneous equations for the coefficients in \(y_{p}\). Then solve for the coefficients to obtain \(y_{p}\). Compare the work you've done with the work required to obtain the same results in Examples \(5.4 .1-5.4 .6 .\) Compare with Example 5.4 .5:$ $$ y^{\prime \prime}-4 y^{\prime}+3 y=e^{3 x}\left(6+8 x+12 x^{2}\right), \quad y_{p}=e^{3 x}\left(A x+B x^{2}+C x^{3}\right) $$

Short answer

In conclusion, we have successfully found the particular solution \(y_p\) for the given second-order non-homogeneous linear differential equation by substituting the suggested form of \(y_p = e^{3x}(Ax + Bx^2 + Cx^3)\), equating coefficients of like terms, and solving for the coefficients A, B, and C. Our final answer is \(y_p = e^{3x}\left(\frac{3}{2}x + \frac{4}{13}x^2\right)\), and this result is consistent with Example 5.4.5, confirming the correctness of our approach.

Step-by-step solution

Compute \(y_p'\) and \(y_p''\)

We first need to calculate the first and second derivatives of the given particular solution, \(y_p = e^{3x}\left(Ax + Bx^2 + Cx^3\right)\). First derivative, \(y_p'\): $$ y_p' = \left(e^{3x}\left(Ax + Bx^2 + Cx^3\right)\right)' = \left(e^{3x}\right)' \left(Ax + Bx^2 + Cx^3\right) + e^{3x}\left(\left(Ax + Bx^2 + Cx^3\right)'\right) $$ $$ y_p' = 3e^{3x}\left(Ax + Bx^2 + Cx^3\right) + e^{3x}\left(A + 2Bx + 3Cx^2\right) $$ Second derivative, \(y_p''\): $$ y_p'' = \left(y_p'\right)' = \left( 3e^{3x}\left(Ax + Bx^2 + Cx^3\right) + e^{3x}\left(A + 2Bx + 3Cx^2\right) \right)' $$ $$ y_p'' = 9e^{3x}\left(Ax + Bx^2 + Cx^3\right) + 6e^{3x}\left(A + 2Bx + 3Cx^2\right) + 3e^{3x}\left(A + 2Bx + 3Cx^2\right)' $$ $$ y_p'' = 9e^{3x}\left(Ax + Bx^2 + Cx^3\right) + 6e^{3x}\left(A + 2Bx + 3Cx^2\right) + 3e^{3x}\left(2B + 6Cx\right) $$

Substitute \(y_p\), \(y_p'\), and \(y_p''\) into the given differential equation

Now, substitute \(y_p\), \(y_p'\), and \(y_p''\) into the given differential equation: $$ y_p'' - 4y_p' + 3y_p = e^{3x}\left(6 + 8x + 12x^2\right) $$ $$ \begin{aligned} &\left(9e^{3x}\left(Ax + Bx^2 + Cx^3\right) + 6e^{3x}\left(A + 2Bx + 3Cx^2\right) + 3e^{3x}\left(2B + 6Cx\right)\right) \\ & - 4\left(3e^{3x}\left(Ax + Bx^2 + Cx^3\right) + e^{3x}\left(A + 2Bx + 3Cx^2\right)\right) \\ & + 3\left(e^{3x}\left(Ax + Bx^2 + Cx^3\right)\right) \\ & = e^{3x}\left(6 + 8x + 12x^2\right) \end{aligned} $$

Simplify and equate coefficients

Next, simplify the equation and equate the coefficients of like terms. $$ \begin{aligned} &2 e^{3x}(Ax + Bx^2 + Cx^3) + 2 e^{3x}(A + 2Bx + 3Cx^2) + 3e^{3x}(2B + 6Cx) \\ &= e^{3x}(6 + 8x + 12x^2) \end{aligned} $$ Since \(e^{3x}\) is never zero, it can be cancelled out: $$ 2(Ax + Bx^2 + Cx^3) + 2(A + 2Bx + 3Cx^2) + 3(2B + 6Cx) = 6 + 8x + 12x^2 $$ Now, expand and group similar terms: $$ (2A + 2A) + (4Bx + 16Bx + 6B) + (8Cx^2 + 12Cx^2 + 18Cx^3) = 6 + 8x + 12x^2 $$ Simplify: $$ (4A) + (26Bx) + (20Cx^2 + 18Cx^3) = 6 + 8x + 12x^2 $$ Now, we can equate coefficients of the like terms: $$ \begin{aligned} 4A &= 6 \\ 26B &= 8 \\ 20C &= 12 \\ 18C &= 0 \end{aligned} $$

Solve the simultaneous equations for A, B, and C

Now, solve the equations to find A, B, and C: $$ \begin{aligned} A &= \frac{6}{4} = \frac{3}{2} \\ B &= \frac{8}{26} = \frac{4}{13} \\ C &= 0 \end{aligned} $$ Finally, substitute A, B, and C back into \(y_p\): $$ y_p = e^{3x}\left(\frac{3}{2}x + \frac{4}{13}x^2 + 0x^3\right) = e^{3x}\left(\frac{3}{2}x + \frac{4}{13}x^2\right) $$ Given the results obtained, we can compare our work with Example 5.4.5 and see that they are consistent, showing that the method used is correct.

Key concepts

Particular Solution

In differential equations, a particular solution is a solution that represents a specific set of conditions or represents the behavior of a function under certain circumstances. For example, when dealing with a non-homogeneous differential equation, a particular solution will satisfy the equation and all initial or boundary conditions that might be specified along the way.

When deriving a particular solution, one commonly uses a method known as "undetermined coefficients." In this approach, we substitute a proposed form of the solution into the differential equation to determine unknown coefficients. Hence, for our exercise, we had a given form of the particular solution, \(y_{p} = e^{3x}(Ax + Bx^2 + Cx^3)\), which is yet to determine the values of \(A\), \(B\), and \(C\) that allow it to satisfy the equation:

\(y^{\prime \prime} - 4y^{\prime} + 3y = e^{3x}(6 + 8x + 12x^2)\).

We solve it by computing derivatives, plugging them into the equation, and equating the coefficients for like powers of \(x\). By doing this, we undertake systematic steps to discover the particular solution that aligns well with the non-homogeneous portion of the equation.

Homogeneous Equations

Homogeneous equations refer to differential equations set to zero on one side. In other words, they have the form:

\(Ly = 0\)

where \(L\) is a linear differential operator.

They are easier to solve because the behavior of the solutions generally resembles exponential functions or polynomial-exponential functions. The term "homogeneous" here implies uniformity and indicates that all terms of the equation involve the function or its derivatives.

These equations often come into play when solving non-homogeneous differential equations by way of the method of undetermined coefficients or variation of parameters. First, a general solution to the corresponding homogeneous equation is found, and then, a particular solution of the non-homogeneous equation is determined. The final solution becomes a combination of these two parts. For example, in our exercise, although it's non-homogeneous, understanding the behavior of its corresponding homogeneous equation aids in forming its particular solution.

Non-homogeneous Differential Equations

Unlike their homogeneous counterparts, non-homogeneous differential equations include a non-zero right-hand side. Hence, they take the form:

\(Ly = f(x)\)

where \(Ly\) is a linear differential operator, and \(f(x)\) is a typical function that varies based upon the specific problem.

Solving non-homogeneous differential equations involves finding two solutions:

• The complementary solution\(y_{c},\) derived from the associated homogeneous equation,\(Ly = 0\).
• The particular solution \(y_{p},\) which is a specific solution that satisfies\(Ly=f(x)\).

For our exercise, the non-homogeneous equation given was:

\( y^{\prime \prime}-4 y^{\prime}+3 y=e^{3 x}(6+8 x+12 x^{2})\).

We focused on finding the particular solution through a method where coefficients corresponding to similar terms on both sides of the equation are equated, ensuring the equation holds true under given conditions.

Simultaneous Equations

Simultaneous equations are a hallmark of solving systems where multiple equations intersect. In the context of differential equations, they arise when one uses methods like the undetermined coefficients to ascertain constants for the particular solution.

In our scenario, after substituting our assumed particular solution and its derivatives into the differential equation, we arrived at simultaneous equations for \(A\), \(B\), and \(C\). By comparing coefficients from both sides of the equation:

• \(4A = 6\)
• \(26B = 8\)
• \(20C = 12\)
• \(18C = 0\)

Solving these allows estimating the actual values needed to devise the correct particular solution. This intersection-solving process involves basic algebra concepts, ensuring clarity and precision in determining unknowns.

This step-by-step solving climbs upon the foundational solution of simpler algebraic systems, showing the integration of several mathematical strata in differential equation solving.