Problem 31 - Advanced Engineering Mathematics

Question

In Problems 31-34, verify that the given two-parameter family of functions is the general solution of the nonhomogeneous differential equation on the indicated interval. $$ \begin{aligned} &y^{\prime \prime}-7 y^{\prime}+10 y=24 e^{x} \\ &y=c_{1} e^{2 x}+c_{2} e^{5 x}+6 e^{x},(-\infty, \infty) \end{aligned} $$

Step-by-Step Solution

Answer

The function \( y = c_1 e^{2x} + c_2 e^{5x} + 6e^x \) is the general solution for \( y'' - 7y' + 10y = 24e^x \) on \((- fty, fty)\).

Step 1: Verify Homogeneous Solution

First, look at the complementary (homogeneous) solution of the differential equation by solving the equation \( y'' - 7y' + 10y = 0 \). The characteristic equation for this is \( r^2 - 7r + 10 = 0 \). Solve for \( r \) to find the roots.

Step 2: Solve Characteristic Equation

The characteristic equation is \( r^2 - 7r + 10 = 0 \). To find the roots, factor the equation: \( (r - 2)(r - 5) = 0 \). Therefore, the roots are \( r = 2 \) and \( r = 5 \). The complementary solution is \( y_c = c_1 e^{2x} + c_2 e^{5x} \).

Step 3: Verify Particular Solution

Now, check the particular solution. The non-homogeneous differential equation is \( y'' - 7y' + 10y = 24e^x \). The proposed particular solution is \( Y_p = 6e^x \). Substitute \( Y_p \) into the left side of the differential equation and verify that it equals \( 24e^x \).

Step 4: Substitute Particular Solution

Compute the first and second derivatives of \( Y_p = 6e^x \): \( Y_p' = 6e^x \) and \( Y_p'' = 6e^x \). Substitute into the differential equation: \( 6e^x - 7(6e^x) + 10(6e^x) = 6e^x - 42e^x + 60e^x = 24e^x \). The left side equals the right side, confirming \( Y_p = 6e^x \) is correct.

Step 5: General Solution Verification

Combine the complementary and particular solutions: \( y = c_1 e^{2x} + c_2 e^{5x} + 6e^x \). Verify this general solution satisfies the original differential equation across \((-fty, fty)\) by checking the linearity and independence of terms, which has been demonstrated in previous steps.

Key Concepts

Complementary Solution

A complementary solution is a vital part of solving a nonhomogeneous differential equation. This solution stems from the homogeneous part of the equation, which is what you get when you set the nonhomogeneous part to zero. In our exercise, the homogeneous equation is \[ y'' - 7y' + 10y = 0. \]By solving this, we find the complementary solution, which is a linear combination of the homogeneous solutions. These homogeneous solutions are found by solving the characteristic equation and are expressed in terms like \(c_1e^{2x}\) and \(c_2e^{5x}\). This solution represents the set of all possible solutions to the homogeneous equation and is crucial for forming the general solution.

Characteristic Equation

The characteristic equation is central to finding the complementary solution of linear differential equations. It is derived from the homogeneous part of the differential equation by assuming a solution of the form \(e^{rx}\) and substituting it into the equation. For the given homogeneous equation \(y'' - 7y' + 10y = 0\), the characteristic equation is \[ r^2 - 7r + 10 = 0. \]By solving this quadratic equation, we determine that the roots are \(r=2\) and \(r=5\). These roots are then used to construct the complementary solution \(y_c = c_1 e^{2x} + c_2 e^{5x}\). The roots lead us to exponential terms that form the backbone of the complementary solution.

Particular Solution

In solving nonhomogeneous differential equations, the particular solution accounts for the nonhomogeneous part. To find it, we need a function that, when substituted into the differential equation, satisfies it completely with the non-zero term on the right side. For our equation, the nonhomogeneous term is \(24e^x\). A reasonable assumption for the particular solution is \(Y_p = 6e^x\). By finding the derivatives and substituting \(Y_p = 6e^x\) into the differential equation, we verify that the left-hand side equals the nonhomogeneous term, \(24e^x\). This confirms that \(Y_p = 6e^x\) is indeed a suitable particular solution.

General Solution

The general solution is the culmination of solving a nonhomogeneous differential equation. It combines both the complementary and particular solutions. The complementary solution, \(y_c = c_1 e^{2x} + c_2 e^{5x}\), addresses the homogeneous part, while the particular solution, \(Y_p = 6e^x\), satisfies the nonhomogeneous aspect. Together, they form the general solution:\[ y = c_1 e^{2x} + c_2 e^{5x} + 6e^x. \]This solution accounts for all possible scenarios by incorporating the constants \(c_1\) and \(c_2\), allowing it to satisfy a wide range of initial or boundary conditions. The general solution fully encapsulates the behavior of the differential equation across its specified interval, \(( -\infty, \infty)\).