Question

(a) Verify that \(y_{1}=e^{2 x}\) and \(y_{2}=e^{5 x}\) are solutions of $$y^{\prime \prime}-7 y^{\prime}+10 y=0$$ on \((-\infty, \infty)\). (b) Verify that if \(c_{1}\) and \(c_{2}\) are arbitrary constants then \(y=c_{1} e^{2 x}+c_{2} e^{5 x}\) is a solution of (A) on \((-\infty, \infty)\) (c) Solve the initial value problem $$y^{\prime \prime}-7 y^{\prime}+10 y=0, \quad y(0)=-1, \quad y^{\prime}(0)=1$$ (d) Solve the initial value problem $$y^{\prime \prime}-7 y^{\prime}+10 y=0, \quad y(0)=k_{0}, \quad y^{\prime}(0)=k_{1}$$

Short answer

- The given functions are \(y_1=e^{2x}\) and \(y_2=e^{5x}\), and the differential equation is \(y''-7y'+10y=0\). 2. Are these functions solutions to the given differential equation? - Yes, both \(y_1=e^{2x}\) and \(y_2=e^{5x}\) are solutions to the given differential equation. 3. What is the general solution of the differential equation? - The general solution is in the form of \(y=c_1e^{2x}+c_2e^{5x}\). 4. What is the specific solution to the initial value problem given \(y(0)=-1\) and \(y'(0)=1\)? - The specific solution is \(y=-2e^{2x}+e^{5x}\). 5. What is the general solution to the initial value problem given \(y(0)=k_0\) and \(y'(0)=k_1\)? - The general solution remains the same as before: \(y=c_1e^{2x}+c_2e^{5x}\).

Step-by-step solution

Verify if \(y_1=e^{2x}\) and \(y_2=e^{5x}\) are solutions of the given differential equation

First, we need to compute the first and second derivatives of both functions. For \(y_1=e^{2x}\): $$y_1'=2e^{2x}, \quad y_1''=4e^{2x}$$ For \(y_2=e^{5x}\): $$y_2'=5e^{5x}, \quad y_2''=25e^{5x}$$ Next, we need to plug these values into the original differential equation to see if it holds true. If it does, then the functions \(y_1\) and \(y_2\) are solutions. For \(y_1\): $$(4e^{2x}) - 7(2e^{2x}) + 10(e^{2x}) = 0$$ For \(y_2\): $$(25e^{5x}) - 7(5e^{5x}) + 10(e^{5x}) = 0$$ Since both of these equalities hold true, we can conclude that \(y_1\) and \(y_2\) are solutions to the given differential equation.

Check if the general solution will be in the form of \(y=c_1e^{2x}+c_2e^{5x}\)

Now, we need to verify if the general solution will be in the form of \(y=c_1e^{2x}+c_2e^{5x}\). To do this, we will compute the first and second derivatives of the general solution and plug them into the differential equation. For \(y=c_1e^{2x}+c_2e^{5x}\): $$y'=2c_1e^{2x}+5c_2e^{5x}, \quad y''=4c_1e^{2x}+25c_2e^{5x}$$ Plug in the derivatives to check if the equation holds true: $$(4c_1e^{2x}+25c_2e^{5x}) - 7(2c_1e^{2x}+5c_2e^{5x}) + 10(c_1e^{2x}+c_2e^{5x}) = 0$$ This equation is satisfied, so we can conclude the general solution is in the form of \(y=c_1e^{2x}+c_2e^{5x}\).

Solve the initial value problem (c)

We need to solve the initial value problem with the given conditions \(y(0)=-1\) and \(y'(0)=1\). Plug in initial conditions to the general solution and its derivative: $$-1 = c_1e^{(2\cdot0)} + c_2e^{(5\cdot0)}$$ $$1 = 2c_1e^{(2\cdot0)} + 5c_2e^{(5\cdot0)}$$ Solve the system of equations: $$-1 = c_1 + c_2$$ $$1 = 2c_1 + 5c_2$$ From the first equation, \(c_1 = -1 - c_2\). Plug this into the second equation and solve for \(c_2\): $$1 = 2(-1 - c_2) + 5c_2$$ $$1 = -2 - 2c_2 + 5c_2$$ $$3 = 3c_2$$ $$c_2 = 1$$ Now, find \(c_1\) using the first equation: $$c_1 = -1 - c_2 = -1 - 1 = -2$$ Thus, the solution to the initial value problem (c) is: $$y = -2e^{2x} + e^{5x}$$

Solve the initial value problem (d)

We need to solve the initial value problem with the given conditions \(y(0)=k_0\) and \(y'(0)=k_1\). Plug in initial conditions to the general solution and its derivative: $$k_0 = c_1e^{(2\cdot0)} + c_2e^{(5\cdot0)}$$ $$k_1 = 2c_1e^{(2\cdot0)} + 5c_2e^{(5\cdot0)}$$ We have the system of equations: $$k_0 = c_1 + c_2$$ $$k_1 = 2c_1 + 5c_2$$ We cannot solve this system of equations for specific values of \(c_1\) and \(c_2\) because \(k_0\) and \(k_1\) are arbitrary constants. Hence, the general solution for the initial value problem (d) remains the same as before: $$y = c_1e^{2x}+c_2e^{5x}$$

Key concepts

Initial Value Problem

An initial value problem (IVP) in the context of differential equations refers to a scenario where one is provided with a differential equation and the value of the function (and possibly its derivatives) at a certain point. This additional information, usually in the form of conditions such as y(0) = a and y'(0) = b, allows for the determination of a unique solution that satisfies both the differential equation and the initial conditions.

For example, in the given exercise, part (c) presents an IVP where the differential equation is y'' - 7y' + 10y = 0, with initial conditions y(0) = -1 and y'(0) = 1. These conditions specify the value of the solution and its derivative at the point x = 0, which is key to finding constants c_1 and c_2 that make the general solution unique to this specific problem. The process requires substituting these initial values into the general solution and solving for the constants to ensure the solution is tailored to the initial conditions provided.

Second-Order Linear Differential Equation

A second-order linear differential equation has the general form y'' + p(x)y' + q(x)y = g(x), where p(x), q(x), and g(x) are functions of x, and the solution y is a function of x as well. The equation in our exercise, y'' - 7y' + 10y = 0, is a homogenous second-order linear differential equation because g(x) = 0, simplifying the process of finding a solution.

For homogenous equations like ours, solutions often involve exponential functions, which provide the fundamental set of solutions to form the general solution. These fundamental solutions, y_1 and y_2, are typically exponential functions with the variable x because exponential functions have the unique property of being proportional to their own derivatives - a characteristic that is useful when solving such equations. Verifying that e^{2x} and e^{5x} are solutions involves checking that they satisfy the differential equation for all x, as demonstrated in the exercise.

Exponential Functions Solutions

Exponential functions are common solutions to constant-coefficient linear differential equations. They have the form e^{rx}, where r is a constant. This is because they are the only functions whose rate of change (derivative) is directly proportional to the function itself, making them a natural fit for such equations.

In the context of the exercise, the functions y_1 = e^{2x} and y_2 = e^{5x} are verified to be solutions to the differential equation. The fact that the combination of these solutions, y = c_1e^{2x} + c_2e^{5x}, also solves the differential equation, reveals a core property of linear differential equations: their solutions can be linearly combined to form new solutions. The constants c_1 and c_2 are determined by the initial conditions, and such a general solution can satisfy an infinite number of initial value problems, as shown in part (d) of the exercise, where k_0 and k_1 represent arbitrary initial conditions.