Problem 13 - A First Course in Differential Equations with Modeling Applications

Question

The given two-parameter family is a solution of the indicated differential equation on the interval \((-\infty, \infty) .\) Determine whether a member of the family can be found that satisfies the boundary conditions. \(y=c_{1} e^{x} \cos x+c_{2} e^{x} \sin x ; \quad y^{\prime \prime}-2 y^{\prime}+2 y=0\) (a) \(y(0)=1, \quad y^{\prime}(\pi)=0\) (b) \(y(0)=1, \quad y(\pi)=-1\) (c) \(y(0)=1, \quad y(\pi / 2)=1\) (d) \(y(0)=0, \quad y(\pi)=0\).

Step-by-Step Solution

Answer

Part (c) has a solution; the others do not.

Step 1: Differentiate the given function

Start by finding the first and second derivatives of the given function, \( y = c_1 e^x \cos x + c_2 e^x \sin x \). Using the product rule, we have:\[ y' = (c_1 e^x \cos x + c_2 e^x \sin x)' = c_1(e^x \cos x \cdot 1 + e^x(-\sin x)) + c_2(e^x \sin x \cdot 1 + e^x\cos x). \]Simplifying gives:\[ y' = c_1 e^x (\cos x - \sin x) + c_2 e^x (\sin x + \cos x) \] which further simplifies to:\[ y' = e^x(c_1 \cos x + c_2 \sin x) \].Now for the second derivative, \( y'' \):\[ y'' = (c_1 e^x (\cos x - \sin x) + c_2 e^x (\sin x + \cos x))' \].Expand and simplify using product rule on each term:

Step 2: Substitute derivatives into the differential equation

Insert \( y, y', y'' \) into the differential equation \( y'' - 2y' + 2y = 0 \). Substitute derived values from step 1,First term, \( y'' \): as derived in previous step.Second term, \( -2y' \):\[ -2(c_1 e^x (\cos x - \sin x) + c_2 e^x (\sin x + \cos x)) \].Final term, \( 2y \):\[ 2(c_1 e^x \cos x + c_2 e^x \sin x) \].After simplifying, check if the resulting expression is zero as expected for solutions.

Step 3: Check boundary condition for part (a)

Use \( y(0)=1 \) and \( y'(\pi)=0 \):Substitute \( x=0 \) into \( y \):\[ y(0) = c_1 e^0 \cos 0 + c_2 e^0 \sin 0 = c_1 \times 1 + c_2 \times 0 = c_1 = 1 \].Now differentiate and evaluate at \( x=\pi \) for \( y'(\pi)=0 \):From derivative expression, \( y'(x) = e^x((c_1+c_2)\cos x + (c_2-c_1)\sin x) \).Evaluate at \( \pi \):\[ c_1 + c_2 \times (-1) = 0 \Rightarrow c_1 = c_2 \].This gives inconsistent solution, thus part (a) has no solution.

Step 4: Check boundary condition for part (b)

For \( y(0) = 1 \) and \( y(\pi) = -1 \):At \( x = 0 \):\[ \Rightarrow c_1 = 1 \].At \( x = \pi \):\[ y(\pi) = e^\pi(-c_1\cos\pi - c_2\sin\pi) =-1 \Rightarrow -c_1 e^\pi = -1 \Rightarrow c_1 = -e^{-\pi} \].Can't be consistent, thus part (b) has no solution.

Step 5: Check boundary condition for part (c)

For \( y(0) = 1 \) and \( y(\pi/2)=1 \):At \( x = 0 \), \( c_1 = 1 \) as before.At \( x = \pi/2 \):\[ y(\pi/2) = e^{\pi/2}(c_1\cos(\pi/2) + c_2\sin(\pi/2)) = e^{\pi/2}(0 + c_2) = 1 \].\[ c_2 = e^{-\pi/2} \].This solution works, therefore part (c) has a solution.

Step 6: Check boundary condition for part (d)

For \( y(0)=0 \) and \( y(\pi)=0 \):At \( x = 0 \):\[ c_1 = 0 \].At \( x = \pi \):\[ y(\pi) = e^\pi (c_1(-1) + c_2\cdot0) = 0 \Rightarrow c_1 = 0 \].Inconsistency arised in previous steps yields no solution for part (d).

Key Concepts

Boundary Conditions

In differential equations, boundary conditions help us find specific solutions within a family of general solutions. Suppose you have a differential equation solved by a function. Boundary conditions will set limitations or specifications for this solution.
Here, we look at constraints applied at different points, such as:

• The value of the function at a specific point. For instance, we want our function to equal a certain value at zero.
• Requirements for the derivative of the function at certain points. This might include specifying that the rate of change is zero at a given location.

These conditions enable us to find a particular instance from a general solution that fits the given criteria.
Understanding this is crucial when seeking the precise solution required for a given scenario.

Two-Parameter Family

A two-parameter family of solutions is essentially a collection of solutions defined by two arbitrary constants. These constants, often denoted as \(c_1\) and \(c_2\), offer flexibility. By altering their values, different solutions can be modeled to fit specific needs or conditions.
In this particular context, the general solution to the differential equation is specified by both \(c_1\) and \(c_2\). These parameters need to be finely adjusted to meet boundary conditions such as \(y(0) = 1\) and \(y'(\pi) = 0\).
The two-parameter nature offers a wide range of potential solutions, providing more options to tailor a solution that meets established criteria between solutions by playing with these two parameters.

Product Rule

The product rule is a fundamental concept in calculus used for differentiating products of two functions. If you have a function \( u(x) \times v(x) \), the derivative is calculated as follows:

• Differentiate \( u(x) \) while leaving \( v(x) \) unchanged.
• Differentiate \( v(x) \) while leaving \( u(x) \) unchanged.
• Sum the two products: \( (u'v + uv') \).

In our exercise, the product rule helps find the first and second derivatives of functions like \( e^x \cos x \) and \( e^x \sin x \).
This is necessary for plugging the derivatives back into the original differential equation, confirming the function's status as a solution.

Derivative Evaluation

Derivative evaluation refers to determining the slope or the rate of change of a function at a specific point. This is important when boundary conditions require information about how the function is changing at particular input values.
Let's say you have a derivative like \( y'(x) \). To check a boundary condition such as \( y'(\pi) = 0 \), you simply plug \( x = \pi \) into your derivative expression. This allows you to validate if the condition is satisfied for particular values of the parameters \( c_1 \) and \( c_2 \).
This evaluation technique helps check whether the family of solutions accommodates any specific member satisfying the established boundary condition criteria.