Question

Find a particular solution satisfying the given conditions. \(y^{\prime \prime}+y^{\prime}-6 y=6, \quad y=1, y^{\prime}=4\) when \(x=0\)

Short answer

The particular solution is \( y = 2e^{2x} - 1 \).

Step-by-step solution

Solve the homogeneous equation

First, solve the homogeneous part of the differential equation: \[ y^{\text{''}} + y^{\text{'}} - 6y = 0 \]The characteristic equation is: \[ r^2 + r - 6 = 0 \]Solve for r to find the roots.

Factor and solve the characteristic equation

The characteristic equation \( r^2 + r - 6 = 0 \) factors to \[ (r + 3)(r - 2) = 0 \]This gives the roots of \[ r = -3 \] and \[ r = 2 \].

Write the general solution to the homogeneous equation

Using the roots from the characteristic equation, the general solution to the homogeneous equation is: \[ y_h = C_1 e^{-3x} + C_2 e^{2x} \]

Solve for the particular solution

Assume a particular solution of the form: \[ y_p = A \]Since the differential equation is \[ y^{\text{''}} + y^{\text{'}} - 6y = 6 \], we substitute \( y_p \) into the equation to find \( A \): \[ 0 + 0 - 6A = 6 \] which simplifies to \[ -6A = 6 \]Thus, \[ A = -1 \].

Combine the solutions

The general solution to the original differential equation is the sum of the homogeneous and particular solutions: \[ y = y_h + y_p = C_1 e^{-3x} + C_2 e^{2x} - 1 \]

Apply initial conditions

Use the initial conditions to solve for \( C_1 \) and \( C_2 \):For \( y(0) = 1 \): \[ 1 = C_1 e^{-3(0)} + C_2 e^{2(0)} - 1 \] which simplifies to \[ 1 = C_1 + C_2 - 1 \] thus, \[ C_1 + C_2 = 2 \]For \( y^{'}(0) = 4 \): \[ y^{'} = -3C_1 e^{-3x} + 2C_2 e^{2x} \] Using \( x = 0 \), \[ 4 = -3C_1 + 2C_2 \]Solve this system of equations.

Solve the system of equations

Solve the system of equations: \[ C_1 + C_2 = 2 \] and \[ -3C_1 + 2C_2 = 4 \]From the first equation, express \( C_2 \): \[ C_2 = 2 - C_1 \] Substitute into the second equation: \[ -3C_1 + 2(2 - C_1) = 4 \], which simplifies to \[ -3C_1 + 4 - 2C_1 = 4 \] Combine like terms: \[ -5C_1 + 4 = 4 \] thus, \[ -5C_1 = 0 \] so, \[ C_1 = 0 \] Substitute \( C_1 = 0 \) back into \( C_2 = 2 - C_1 \) to find \( C_2 = 2 \)

Write the final solution

Using the values of \( C_1 \) and \( C_2 \): \[ y = 0 \times e^{-3x} + 2 \times e^{2x} - 1 \]Thus, the particular solution is: \[ y = 2e^{2x} - 1 \]

Key concepts

Particular Solution

In differential equations, a particular solution is a specific solution that satisfies the non-homogeneous equation under given conditions. The general form of a non-homogeneous differential equation includes a term that is not a function of the unknown variable and its derivatives. For example, given the equation: \[ y^{\text{''}} + y^{\text{'}} - 6y = 6 \] A particular solution aims to find a function, specific to the equation’s non-homogeneous part, in this case, the term '6'. Typically, to find this, we assume a solutions that fits the non-homogeneous term's form. In our example, we assume: \[ y_p = A \] and solve for 'A' by substituting into the differential equation, yielding: \[ 0 + 0 - 6A = 6 \] This simplifies to: \[ A = -1 \] Thus, the particular solution is \[ y_p = -1 \]

Initial Conditions

Initial conditions are specific values given for the function and its derivatives at a particular point, often used to find the constants in the general solution. In this problem, the initial conditions are:

• \( y(0) = 1 \)
• \( y^{'}(0) = 4 \)

By applying these conditions, we can solve for the constants. For the homogeneous solution: \[ y_h = C_1 e^{-3x} + C_2 e^{2x} \] Combine both the homogeneous and particular solutions to form: \[ y = y_h + y_p = C_1 e^{-3x} + C_2 e^{2x} -1 \] Using the initial conditions, we can form two equations: \[ 1 = C_1 + C_2 - 1 \] which simplifies to: \( C_1 + C_2 = 2 \), and: \[ y^{'} = -3C_1e^{-3x} + 2C_2e^{2x} \] Substituting \( x = 0 \): \[ 4 = -3C_1 + 2C_2 \] Solve these equations to find the values of the constants.”

Homogeneous Equation

A homogeneous equation is a differential equation in which the non-homogeneous term is zero. For example, from our main equation: \[ y^{''} + y^{'} - 6y = 6 \] The corresponding homogeneous equation is obtained by setting the right-hand side to zero: \[ y^{''} + y^{'} - 6y = 0 \] The solution to the homogeneous equation is crucial, as it forms the basis for the general solution of the non-homogeneous equation. By solving the homogeneous part, we find the solution: \[ y_h = C_1 e^{-3x} + C_2 e^{2x} \] where \( C_1 \) and \( C_2 \) are constants determined by initial conditions.”

Characteristic Equation

The characteristic equation is derived from the homogeneous differential equation and is used to find its solution. For the differential equation: \[ y^{''} + y^{'} - 6y = 0 \] We assume a solution of the form: \[ y = e^{rx} \] Substituting this into the equation gives us the characteristic equation: \[ r^2 + r - 6 = 0 \] This quadratic equation is solved to find the values of 'r', which gives us the roots. Factoring, we get: \[ (r + 3)(r - 2) = 0 \] Therefore, the roots are:

• \( r = -3 \)
• \( r = 2 \)

With these roots, we form the general solution of the homogeneous equation: \[ y_h = C_1 e^{-3x} + C_2 e^{2x} \] Learning to derive and solve the characteristic equation is key for solving higher-order differential equations.