Question

In Exercises \(24-29\) use the principle of superposition to find a particular solution. $$ y^{\prime \prime}-2 y^{\prime}+2 y=e^{x}(1+x)+e^{-x}\left(2-8 x+5 x^{2}\right) $$

Short answer

Question: Find the particular solution of the following second-order non-homogeneous linear differential equation using the principle of superposition: \(y''(x) - 2y'(x) + 2y(x) = e^{x}(1+x)+e^{-x}(2-8x+5x^{2})\). Answer: The particular solution to the given differential equation is: \(Y(x) = \left(\frac{1}{3} + \frac{2}{9}x\right)e^{x} + (-x^2 + x - 2)e^{-x}\).

Step-by-step solution

Find the particular solution for the first part

To find the particular solution for the first part \(e^{x}(1+x)\), we assume a solution of the form: $$ Y_1(x) = (A + Bx)e^x $$ Now differentiate \(Y_1(x)\) two times and plug these into the given differential equation: $$ Y_1'(x) = (A + (A + B)x)e^{x} + Be^{x} $$ $$ Y_1''(x) = (A + 2A + (A + 3B)x)e^{x} + 2Be^{x} $$ Now substitute \(Y_1(x)\), \(Y_1'(x)\), and \(Y_1''(x)\) into the given equation and equate the coefficients of the like terms: $$ (A + 2A + (A + 3B)x)e^{x} + 2Be^{x} - 2(A + (A + B)x)e^{x} + 2Be^{x} = e^{x}(1 + x) $$ From this, we get: $$ A + 2A = 1 \;\Rightarrow\; A = \frac{1}{3} $$ $$ A + 3B = 1 \;\Rightarrow\; B = \frac{2}{9} $$ So, the particular solution for the first part is: $$ Y_1(x) = \left(\frac{1}{3} + \frac{2}{9}x\right)e^{x} $$

Find the particular solution for the second part

To find the particular solution for the second part \(e^{-x}\left(2-8x+5x^2\right)\), we assume a solution of the form: $$ Y_2(x) = (Ax^2 + Bx + C)e^{-x} $$ Now differentiate \(Y_2(x)\) two times and plug these into the given differential equation: $$ Y_2'(x) = e^{-x}(-Ax^2 + (2A-B)x + (A-2B+C)) $$ $$ Y_2''(x) = e^{-x}(Ax^2 - (4A-2B)x + (4A-4B+C)) $$ Now substitute \(Y_2(x)\), \(Y_2'(x)\), and \(Y_2''(x)\) into the given equation and equate the coefficients of the like terms: $$ (Ax^2 - (4A-2B)x + (4A-4B+C))e^{-x} - 2(-Ax^2 + (2A-B)x + (A-2B+C))e^{-x} + 2(Ax^2 + Bx + C)e^{-x} = e^{-x}(2-8x+5x^2) $$ From this, we get: $$ -5A = 5 \;\Rightarrow\; A = -1 $$ $$ 8B - 2A = 8 \;\Rightarrow\; B = 1 $$ $$ 4A - 4B + C = 2 \;\Rightarrow\; C = -2 $$ So, the particular solution for the second part is: $$ Y_2(x) = (-x^2 + x - 2)e^{-x} $$

Combine the particular solutions

Finally, to obtain the particular solution to the entire non-homogeneous equation, we can use the principle of superposition and add the individual particular solutions \(Y_1(x)\) and \(Y_2(x)\): $$ Y(x) = Y_1(x) + Y_2(x) = \left(\frac{1}{3} + \frac{2}{9}x\right)e^{x} + (-x^2 + x - 2)e^{-x} $$ Therefore, the particular solution to the given differential equation is: $$ Y(x) = \left(\frac{1}{3} + \frac{2}{9}x\right)e^{x} + (-x^2 + x - 2)e^{-x} $$

Key concepts

Superposition Principle

The Superposition Principle is a fundamental concept in differential equations used to solve linear systems. It states that if you have a linear differential equation and you need to find its particular solution, you can add up the solutions of individual parts of the equation to form a complete solution.

This principle is exceptionally useful when dealing with complex differential equations composed of several functions. For instance, given a function made up of multiple terms, you can address each term separately. Each of these terms will have its own particular solution.

Once you find these smaller solutions, you sum them together to get the entire particular solution for the original equation.

The principle of superposition simplifies the process by breaking the problem into manageable pieces, which are typically easier to solve. Especially in cases where the differential equation is non-homogeneous, as seen in this exercise, this principle enables a step-by-step solution by handling each component separately.

Particular Solution

A particular solution in differential equations refers to a specific solution that satisfies not only the equation itself but also additional conditions or particular scenarios given in the problem. Each different part of a non-homogeneous differential equation can have its own particular solution.

In the exercise, two functions needed particular solutions: one is based on a term with an exponential function without a coefficient of \(-x\), and the other with \(-x\). For each term, we assume a distinct form for the solution ( e.g.) \((A + Bx)e^x\) or \((Ax^2 + Bx + C)e^{-x}\).

These assumed forms are then differentiated and substituted back into the differential equation. By equating coefficients, we can solve for unknown parameters like \(A, B,\) or \(C\), allowing us to find the particular solution for each part of the equation.

It's crucial to solve each segment efficiently, in order to piece together the entire particular solution using the superposition principle.

Homogeneous Equation

A homogeneous differential equation is one in which the right-hand side of the equation is zero. The form of a homogeneous differential equation is generally given as: \( \ a_n(x)y^{(n)} + a_{n-1}(x)y^{(n-1)} + ... + a_1(x)y' + a_0(x)y = 0 \)
In such equations, solutions often take the form of combinations or constants.The key feature of homogeneous equations is that their solutions form a vector space, so any linear combination of solutions is also a solution. The complementary function, obtained from the homogeneous equation, is combined with the particular solution of a non-homogeneous equation to find the complete solution.

In the context of the step-by-step solution provided, the homogeneous part isn't explicitly detailed but is understood as the complementary function necessary to achieve the general solution when combined with the particular solution derived for the non-zero right-hand side of the non-homogeneous equation.

Non-homogeneous Equation

Unlike homogeneous equations, non-homogeneous differential equations have a non-zero right-hand side, indicated by a function that is not part of the derivatives or variable terms of the equation itself.

The general form of a non-homogeneous equation might look something like:\( a_n(x)y^{(n)} + a_{n-1}(x)y^{(n-1)} + ... + a_1(x)y' + a_0(x)y = g(x) \)Where \(g(x)\) is a non-zero function. The presence of a non-zero \(g(x)\) requires finding a particular solution in addition to the complementary solution derived from the associated homogeneous equation.

In our exercise, the non-homogeneous equation presented has multiple terms on the right-hand side like \(e^{x}(1+x)\) and \(e^{-x}(2-8x+5x^2)\). These components make the equation non-homogeneous, prompting us to find a distinct particular solution for each term, then combine them with any solutions from the associated homogeneous equation to achieve a complete solution. By handling each term on the right-hand side independently, solving becomes more organized and achievable.