Question

Solve the initial value problem. $$ y^{\prime \prime}-7 y^{\prime}+6 y=-e^{x}(17 \cos x-7 \sin x), \quad y(0)=4, y^{\prime}(0)=2 $$

Short answer

Based on the step-by-step solution, the final solution for the given initial value problem is $$y(x) = -\frac{3}{5} e^{x} + \frac{16}{5} e^{6x} + e^{x}(\cos x - \frac{7}{17} \sin x).$$

Step-by-step solution

Solve the Homogeneous Problem

First, we need to solve the corresponding homogeneous problem: $$ y^{\prime \prime}-7 y^{\prime}+6 y = 0 $$ To do this, we first look for a general solution in the form of \(y(x) = e^{rx}\). Substituting this into the equation, we get: $$ r^2 e^{rx} - 7r e^{rx} + 6 e^{rx} = 0 $$ Now, we can factor out the \(e^{rx}\) term, since it is never 0: $$ e^{rx}(r^2 - 7r + 6) = 0 $$ This leaves us with the quadratic equation \(r^2 - 7r + 6 = 0\). To find its roots, we can factor it as \((r-1)(r-6)=0\). This gives us \(r_1 = 1\) and \(r_2 = 6\). Therefore, the general solution to the homogeneous problem is: $$ y_h(x) = C_1 e^{x} + C_2 e^{6x} $$

Find a Particular Solution

Now, we need to find a particular solution for the non-homogeneous equation: $$ y^{\prime \prime}-7 y^{\prime}+6 y = -e^{x}(17 \cos x-7 \sin x) $$ We can guess a particular solution of the form: $$ y_p(x) = e^{x}(A \cos x + B \sin x) $$ Now, we find the first and second derivatives of \(y_p(x)\) and substitute them into the non-homogeneous equation: $$ y_p^{\prime}(x) = e^{x}(A \sin x - A \cos x + B \cos x + B \sin x) $$ and $$ y_p^{\prime\prime}(x) = e^{x}(2A \sin x - 2A \cos x - 2B \cos x - 2B \sin x) $$ Substitute \(y_p(x), y_p^{\prime}(x),\) and \(y_p^{\prime\prime}(x)\) into the given equation and simplify: $$ e^{x}(17 \cos x - 7 \sin x) = -e^{x}(17 \cos x - 7 \sin x) $$ From this, we can find the values of \(A\) and \(B\): $$ A = 1, \quad B = -\frac{7}{17} $$ So, the particular solution is: $$ y_p(x) = e^{x}(\cos x - \frac{7}{17} \sin x) $$

Determine the Constants using Initial Conditions

Now, we have the general solution: $$ y(x) = y_h(x) + y_p(x) = C_1 e^{x} + C_2 e^{6x} + e^{x}(\cos x - \frac{7}{17} \sin x) $$ Now, using the initial conditions \(y(0) = 4\) and \(y^{\prime}(0) = 2\), we solve for \(C_1\) and \(C_2\): $$ 4 = y(0) = C_1 e^{0} + C_2 e^{0} + e^{0}(\cos 0 - \frac{7}{17} \sin 0) \implies C_1 + C_2 + 1 = 4 $$ and $$ 2 = y^{\prime}(0) = C_1 e^{0} - C_1 e^{0} + 6 C_2 e^{0} + e^{0}(0 - \frac{7}{17}(0)) \implies 5C_1 + 5 = 2 $$ Solving the system of equations, we obtain \(C_1 = -\frac{3}{5}\) and \(C_2 = \frac{16}{5}\).

Final Solution

Finally, we plug the values of \(C_1\) and \(C_2\) back into the general solution to get the final solution for the initial value problem: $$ y(x) = -\frac{3}{5} e^{x} + \frac{16}{5} e^{6x} + e^{x}(\cos x - \frac{7}{17} \sin x) $$

Key concepts

Homogeneous Differential Equations

The first step in solving second-order linear differential equations often involves addressing the corresponding homogeneous equation. A homogeneous differential equation refers to a differential equation where the right-hand side is zero. For example, the homogeneous form of our given problem is:
\[ y'' - 7y' + 6y = 0 \]
The goal is to find solutions that satisfy this equation. We begin by hypothesizing that the solution can be expressed as an exponential function \( y(x) = e^{rx} \). This transforms the differential equation into a characteristic equation, a form of a quadratic equation:

• \( r^2 e^{rx} - 7r e^{rx} + 6 e^{rx} = 0 \)

Since \( e^{rx} \) is never zero, we simplify to find the roots of the corresponding characteristic polynomial:
\[ r^2 - 7r + 6 = 0 \]
By solving this quadratic equation, we determine the roots \( r_1 = 1 \) and \( r_2 = 6 \). The general solution to the homogeneous differential equation is then given by the combination:
\[ y_h(x) = C_1 e^{x} + C_2 e^{6x} \]where \( C_1 \) and \( C_2 \) are constants to be determined later.

Particular Solution

Moving on from the homogeneous equation, we seek a particular solution to the non-homogeneous differential equation. This involves tackling the equation:
\[ y'' - 7y' + 6y = -e^{x}(17\cos x - 7\sin x) \]
For non-homogeneous equations, the method of undetermined coefficients is a common strategy. We hypothesize a solution resembling the right-hand side of the equation, often multiplied by a probable factor. In this scenario, proposing:
\[ y_p(x) = e^{x}(A \cos x + B \sin x) \]
To determine the specific values for \( A \) and \( B \):

• Calculate the derivatives \( y_p'(x) \) and \( y_p''(x) \).
• Substitute these expressions back into the differential equation.
• Equate coefficients of like terms (\( \cos x \) and \( \sin x \)).

Through this process, we deduce:

• \( A = 1 \)
• \( B = -\frac{7}{17} \)

Thus, the particular solution is:
\[ y_p(x) = e^{x}(\cos x - \frac{7}{17} \sin x) \]combining with the homogeneous solution, we will form the complete solution.

Initial Value Problem

An initial value problem (IVP) stipulates specific conditions at the start of the solution process, providing us with the means to determine unknown constants in differential equations. Our problem includes the conditions:

• \( y(0) = 4 \)
• \( y'(0) = 2 \)

To incorporate these, we combine the homogeneous and particular solutions:
\[ y(x) = C_1 e^{x} + C_2 e^{6x} + e^{x}(\cos x - \frac{7}{17} \sin x) \]
Substitute \( x = 0 \) into the general solution and its derivative to solve for \( C_1 \) and \( C_2 \). This becomes a simple algebra problem:

• Start with calculating \( y(0) \).
• Differentiate \( y(x) \) and substitute into \( y'(0) \).

Solve these equations simultaneously to obtain \( C_1 = -\frac{3}{5} \) and \( C_2 = \frac{16}{5} \). Having these constants allows us to express the solution satisfying the initial conditions provided.

Quadratic Equation Solutions

A crucial part of solving homogeneous differential equations involves solving a quadratic equation derived from the characteristic polynomial. A quadratic equation generally looks like:
\[ ax^2 + bx + c = 0 \]
For the homogeneous problem:
\[ r^2 - 7r + 6 = 0 \]
This equation represents a simplified form, since the coefficient \( a=1 \). Solving such equations can be easily achieved through:

• Factoring: Splitting into products, such as \((r-1)(r-6) = 0\).
• Using the quadratic formula: \( r = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \).

For this instance, the roots, obtained by factoring, are \( r_1 = 1 \) and \( r_2 = 6 \), which guide us in forming the general homogeneous solution:
\[ y_h(x) = C_1 e^{x} + C_2 e^{6x} \]Understanding how to solve these quadratic equations is fundamental for addressing more complex differential equations and ensures accurate solutions.