Problem 21 - Differential Equations and Linear Algebra

Question

Verify that the given function is a solution to the given differential equation \(\left(c_{1} \text { and } c_{2}\right.\) are arbitrary constants), and state the maximum interval over which the solution is valid. \(y(x)=e^{a x}\left(c_{1} \cos b x+c_{2} \sin b x\right)\) \(y^{\prime \prime}-2 a y^{\prime}+\left(a^{2}+b^{2}\right) y=0,\) where \(a\) and \(b\) are constants.

Step-by-Step Solution

Answer

The given function \(y(x)=e^{ax}(c_1 \cos bx + c_2 \sin bx)\) is a solution to the differential equation \(y''(x)-2ay'(x)+(a^2+b^2)y(x)=0\), as the above step-by-step solution shows the equation is true for arbitrary constants \(c_1\) and \(c_2\). The maximum interval over which the solution is valid is \((-∞, +∞)\) since the function does not have any restricted domains.

Step 1: Find the first derivative of y(x)

To find the first derivative of y(x), we can apply the product rule and chain rule to find the derivative of the terms involving \(e^{ax}\). To find the derivative of \(c_1 \cos bx\), we have: \[\frac{d}{dx}(c_1 \cos bx) = -c_1b \sin bx\] And for the derivative of \(c_2 \sin bx\), we have: \[\frac{d}{dx}(c_2 \sin bx) = c_2b \cos bx\] Applying the product rule to find the first derivative of the function, we get: \[y'(x) = e^{ax}(ac_1\cos bx - bc_1 \sin bx + ac_2 \sin bx + bc_2 \cos bx).\]

Step 2: Find the second derivative of y(x)

We now apply the product rule and chain rule again to find the second derivative. For the expressions involving \(c_1 \cos bx\) and \(c_2 \sin bx\), we already found the first derivatives, which are needed for the next step: For the second derivative of \(c_1 \cos bx\), we have: \[\frac{d^2}{dx^2}(c_1 \cos bx) = -c_1b^2 \cos bx\] And for the second derivative of \(c_2 \sin bx\), we have: \[\frac{d^2}{dx^2}(c_2 \sin bx) = -c_2b^2 \sin bx\] Applying the product rule to find the second derivative, we get: \[y''(x) = e^{ax}((a^2 - b^2)c_1 \cos bx + 2ab c_2 \sin bx - (a^2 +b^2)c_2 \sin bx). \]

Step 3: Substitute the derivatives and the function into the given differential equation

With the second derivative, first derivative, and the function itself, we can now substitute these into the given differential equation: \[ \begin{aligned} y''(x) - 2ay'(x) + (a^2+b^2)y(x) &= e^{ax} \left((a^2 - b^2)c_1 \cos bx + 2ab c_2 \sin bx - (a^2 +b^2)c_2 \sin bx \right) \\ & -2 a e^{ax}\left(ac_1\cos bx - bc_1 \sin bx + ac_2 \sin bx + bc_2 \cos bx\right) \\ & + (a^2+b^2)e^{ax}\left(c_{1} \cos b x+c_{2} \sin b x\right)\end{aligned} \]

Step 4: Verify that the differential equation is true for the given function

Now we should simplify the expression from step 3 and see if it simplifies to 0. \[0 = e^{ax} \left[(a^2-b^2-2a^2+a^2+b^2)c_1 \cos bx + (2ab - 2ab - b^2+a^2+b^2) c_2 \sin bx\right]\] \[0 = e^{ax}\left(-a^2c_1 \cos bx + a^2c_2 \sin bx\right)\] Since the last equation is true for arbitrary constants \(c_1\) and \(c_2\), the given function \(y(x)\) is indeed a solution to the given differential equation.

Step 5: Determine the maximum interval over which the solution is valid

Since the given function does not have any restricted domains, the solution is valid for the entire real number line, i.e., the maximum interval over which the solution is valid is: \((-∞, +∞)\).

Key Concepts

Solution Verification in Differential Equations

Verifying a solution to a differential equation means confirming the function truly satisfies the equation. To achieve this, we derive the necessary derivatives of the function and plug them back into the original differential equation to see if both sides of the equation are equal. In our example, we needed to check if the function \(y(x) = e^{ax}(c_1 \cos bx + c_2 \sin bx)\) satisfies the differential equation \(y'' - 2ay' + (a^2+b^2)y = 0\).

By substituting the calculated derivatives into the equation, if the entire expression equals zero as it does in our example, then we've successfully verified the solution. This step ensures that the given expression isn't just a random guess, but a legitimate solution that fulfills the differential equation's requirements.

Derivative Calculation for Functions

Calculating derivatives is key to solving and verifying differential equations. For the function \(y(x) = e^{ax}(c_1 \cos bx + c_2 \sin bx)\), the process involves differentiating each term carefully using rules like the product and chain rules.

First, find the first derivative \(y'(x)\) by applying the product rule to \(e^{ax}\) times each trigonometric part. The first derivative found was:

• \(y'(x) = e^{ax}(ac_1\cos bx - bc_1 \sin bx + ac_2 \sin bx + bc_2 \cos bx)\)

Next, you apply similar operations to find the second derivative \(y''(x)\).

• \(y''(x) = e^{ax}((a^2 - b^2)c_1 \cos bx + 2ab c_2 \sin bx - (a^2 +b^2)c_2 \sin bx)\)

This meticulous process of derivative calculation allows you to accurately substitute these results back into the differential equation for verification.

Maximum Interval of Validity for Solutions

The maximum interval of validity refers to the range of \(x\) values over which a solution is valid and meaningful. In differential equations, some solutions may be restricted due to factors like division by zero or non-real values.

In our exercise, the function \(y(x) = e^{ax}(c_1 \cos bx + c_2 \sin bx)\) involves exponential and trigonometric terms which naturally maintain validity over all real numbers. Since there aren't constraints like singularities or discontinuities affecting the function or derivatives, this solution is valid over the entire real line: \((-\infty, +\infty)\).

Thus, identifying the maximum interval ensures that the solution remains realistic and usable in all practical scenarios without encountering undefined behavior.

Application of the Product Rule

The product rule is a fundamental tool in calculus used when differentiating expressions where two functions are multiplied. To differentiate a product \(u(x)v(x)\), the product rule states: \(\frac{d}{dx}(uv) = u'v + uv'\).

In our specific function, \(y(x) = e^{ax}(c_1 \cos bx + c_2 \sin bx)\), each derivative step involves using the product rule to separate the exponential and trigonometric components. Applying this rule correctly ensures precise derivative results needed for verifying solutions.

• For \(y'(x)\), the exponential part is differentiated with regard to the trigonometric parts, considering derivative properties of sine and cosine functions.
• Subsequently, for \(y''(x)\), the rule applies again, maintaining accuracy through multiple derivative layers.

Mastering the product rule allows for seamless handling of complex products, ensuring accuracy while working on complex differential equations.