Problem 3 - Advanced Engineering Mathematics

Question

Solve the given differential equation by undetermined coefficients. \(y^{\prime \prime}-10 y^{\prime}+25 y=30 x+3\)

Step-by-Step Solution

Answer

The solution is \(y = C_1e^{5x} + C_2xe^{5x} + \frac{6}{5}x + \frac{9}{25}\).

Step 1: Identify the Type of Differential Equation

The given differential equation is \(y^{\prime \prime} - 10y^{\prime} + 25y = 30x + 3\). It is a second-order linear non-homogeneous differential equation with constant coefficients.

Step 2: Solve the Homogeneous Equation

Start by solving the associated homogeneous equation: \(y^{\prime \prime} - 10y^{\prime} + 25y = 0\). The characteristic equation is \(r^2 - 10r + 25 = 0\). Solve for \(r\) using the quadratic formula or by factoring: \((r - 5)^2 = 0\). Thus, \(r = 5\) is a double root, and the complementary solution is \(y_c = C_1e^{5x} + C_2xe^{5x}\).

Step 3: Form of the Particular Solution

Since the non-homogeneous part is a polynomial, i.e., \(30x + 3\), assume a particular solution of the form \(y_p = Ax + B\).

Step 4: Determine Coefficients of the Particular Solution

Substitute \(y_p = Ax + B\), \(y_p^{\prime} = A\), \(y_p^{\prime \prime} = 0\) into the differential equation: \(0 - 10A + 25(Ax + B) = 30x + 3\). Simplify and compare coefficients to solve for \(A\) and \(B\): For \(x\) term: \(25A = 30\) implies \(A = \frac{6}{5}\).For constant term: \(-10A + 25B = 3\) implies \(B = \frac{9}{25}\). Thus, \(y_p = \frac{6}{5}x + \frac{9}{25}\).

Step 5: Write the General Solution

Combine the complementary and particular solutions: \(y = y_c + y_p = C_1e^{5x} + C_2xe^{5x} + \frac{6}{5}x + \frac{9}{25}\).

Step 6: Conclusion

Thus, the solution to the differential equation is \(y = C_1e^{5x} + C_2xe^{5x} + \frac{6}{5}x + \frac{9}{25}\).

Key Concepts

Second-Order Differential Equation

A second-order differential equation is one of the most fascinating areas of calculus and mathematics. It involves derivatives up to the second order (or the second derivative) of an unknown function.
The general form of a second-order differential equation is:

• \(a y^{\prime\prime} + b y^{\prime} + c y = f(x)\)

where \(a, b, c\) are constants and \(f(x)\) is a given function. In our specific exercise, we see the equation \(y^{\prime \prime} - 10y^{\prime} + 25y = 30x + 3\).
Studying these equations helps in understanding many physical phenomena, such as motion, electricity, and more.

Linear Non-Homogeneous Differential Equation

A linear non-homogeneous differential equation is characterized by its form where the terms are either functions of the independent variable, its derivatives, or a constant. The key part is the non-homogeneity, which indicates the presence of an external term separated from the ordinary linear part.
Such equations are generally represented as:

• \(a y^{\prime\prime} + b y^{\prime} + c y = g(x)\)

where \(g(x)\) is a non-zero function. In our example with \(y^{\prime \prime} - 10y^{\prime} + 25y = 30x + 3\), the non-homogeneous part is the polynomial \(30x + 3\).
Learning to solve these takes into account both the structure and the independent non-homogeneous component.

Complementary Solution

The complementary solution is crucial for solving linear differential equations. It relates to the solutions of the homogeneous part of the equation. By setting the non-homogeneous component to zero, you derive a homogeneous equation. Solving this gives us the complementary solution.
For \(y^{\prime \prime} - 10y^{\prime} + 25y = 0\), the characteristic equation is derived as:

• \((r - 5)^2 = 0\), solved to find \(r = 5\) (a double root).

The complementary solution thus becomes:

• \(y_c = C_1e^{5x} + C_2xe^{5x}\)

The constants \(C_1\) and \(C_2\) allow flexibility depending on the initial conditions of the specific problem being solved.

Particular Solution

The particular solution addresses the non-homogeneous part of the differential equation. It is a specific solution that satisfies not only the structure of the equation but also the non-homogeneous term.
Assuming a particular solution form (based on the non-homogeneous term) is essential. For our equation with a polynomial \(30x + 3\), we assume:

• \(y_p = Ax + B\)

Plugging \(y_p\) into the differential equation allows one to solve for the coefficients \(A\) and \(B\) by matching terms.
Here,

• \(A\) is found as \(\frac{6}{5}\) and \(B\) as \(\frac{9}{25}\), leading to the particular solution: \(y_p = \frac{6}{5}x + \frac{9}{25}\).

The particular and complementary solutions together provide the overall solution to the differential equation.