Question

Finding a Particular Solution In Exercises \(35-40\) , verify that the general solution satisfies the differential equation. Then find the particular solution that satisfies the initial condition(s). $$ \begin{array}{l}{y=C e^{-2 x}} \\ {y^{\prime}+2 y=0} \\ {y=3 \text { when } x=0}\end{array} $$

Short answer

The particular solution that satisfies the given initial condition is \(y = 3e^{-2x}\).

Step-by-step solution

Verify the General Solution

Differentiate the function \(y = Ce^{-2x}\) using the chain rule to get \(y' = -2Ce^{-2x}\). Substituting the values of \(y\) and \(y'\) into the differential equation \(y' + 2y = 0\), we get \(-2Ce^{-2x} + 2Ce^{-2x} = 0\), which simplifies to \(0 = 0\). Therefore, the general solution satisfies the differential equation.

Find the Particular Solution

Substitute the initial conditions \(x = 0\) and \(y = 3\) into the general solution \(y = Ce^{-2x}\) to get \(3 = Ce^{0}\) or \(3 = C\). Therefore, the particular solution is \(y = 3e^{-2x}\).

Key concepts

Verifying General Solutions

When presented with a differential equation, a general solution may be given that incorporates a constant, typically represented as C. This general solution is intended to represent a family of functions that could potentially satisfy the original differential equation under various conditions. Verifying that the general solution meets the differential equation's requirements is a crucial check to ensure its validity.

The verification process involves differentiating the general solution according to the rules of differentiation, here the chain rule comes particularly handy. Once we have the derivative, it is substituted back into the original differential equation alongside the general solution to inspect the equality. A successful verification will see the differential equation reduced to an identity such as 0=0, indicating that the general solution holds for any value of C.

To verify the provided general solution in the textbook exercise, we follow this exact process of differentiation and substitution, confirming its correctness as it simplifies properly to an identity.

Initial Conditions

Differential equations often come with initial conditions, which are specific values given for the independent and dependent variables at a certain point. These conditions serve to identify a unique, particular solution from the infinite family of solutions provided by the general solution.

Applying initial conditions is straightforward: we replace the independent variable and the solution function in the general equation with the given values. This typically results in an equation with the constant C as the only unknown. By solving for C, we determine the specific value needed to meet the initial condition, thus obtaining the particular solution to the differential equation.

For instance, the initial condition in the exercise specifies the function's value when the independent variable is zero (x=0). Substituting these values into the general solution gives us the particular constant C, which we can then use to express the unique solution tailored to the initial condition.

Chain Rule Differentiation

The chain rule is a fundamental differentiation rule in calculus that is utilized when dealing with composite functions - functions of functions. The rule can be succinctly expressed as 'the derivative of the outside function with respect to the inside function, times the derivative of the inside function with respect to the variable'.

In the context of our differential equations, when the solution y is expressed as a function of another function, such as an exponential or trigonometric function, the chain rule allows us to find the derivative y' with respect to the independent variable x. This derivative is then used in the process of verifying the general solution and, if initial conditions are provided, determining the particular solution.

In the textbook solution provided, applying the chain rule to differentiate the exponential function within the general solution leads to finding the derivative y', which is then used to show that it satisfies the differential equation.

Solving Differential Equations

Solving differential equations is about finding a function or a set of functions that satisfy the equation. The process generally starts by integrating the function, if possible, to obtain the general solution, which includes an arbitrary constant C. From there, the solution can be further refined by applying specific conditions known as initial or boundary conditions.

There are numerous methods to solve different types of differential equations—separation of variables, integrating factors, or transformation of variables, to name just a few. Each method applies to particular kinds of differential equations and exploits their structure to simplify and solve them.

The exercise provided showcases an example of a first order linear homogeneous differential equation. The process of solving such an equation involves deriving the general solution and fine-tuning it with the given initial conditions, leading up directly to the particular solution. It demonstrates a comprehensive approach to solving differential equations, exemplifying both the general solution derivation and the application of initial conditions.