Problem 1 - A First Course in Differential Equations with Modeling Applications

Question

Verify that \(y_{1}\) and \(y_{2}\) are solutions of the given differential equation but that \(y=c_{1} y_{1}+c_{2} y_{2}\) is, in general, not a solution. $$\left(y^{\prime \prime}\right)^{2}=y^{2} ; \quad y_{1}=e^{x}, y_{2}=\cos x$$

Step-by-Step Solution

Answer

The functions \(y_1 = e^x\) and \(y_2 = \cos x\) are solutions, but the combination \(y = c_1y_1 + c_2y_2\) generally is not.

Step 1: Differentiate y1 and y2

To verify the solutions, first differentiate the functions. The first derivative of \(y_1 = e^x\) is \(y_1' = e^x\), and the second derivative is \(y_1'' = e^x\). For \(y_2 = \cos x\), the first derivative is \(y_2' = -\sin x\), and the second derivative is \(y_2'' = -\cos x\).

Step 2: Verify y1 as a solution

Substitute \(y_1 = e^x\) into the differential equation \((y'')^2 = y^2\). This gives \((e^x)^2 = (e^x)^2\), which is \(e^{2x} = e^{2x}\), confirming that \(y_1\) is a solution.

Step 3: Verify y2 as a solution

Substitute \(y_2 = \cos x\) into the differential equation \((y'')^2 = y^2\). This gives \((-\cos x)^2 = (\cos x)^2\), or \(\cos^2 x = \cos^2 x\), confirming that \(y_2\) is a solution.

Step 4: Substitute the general solution y

Consider the general solution \(y = c_1 e^x + c_2 \cos x\). Differentiate twice to find \(y'' = c_1 e^x - c_2 \cos x\).

Step 5: Check the general solution

Substitute \(y'' = c_1 e^x - c_2 \cos x\) into \((y'')^2 = y^2\). Expanding yields \((c_1 e^x - c_2 \cos x)^2 eq (c_1 e^x + c_2 \cos x)^2\) unless \(c_1 = 0\) or \(c_2 = 0\). Thus, \(y\) is generally not a solution.

Key Concepts

Solutions of Differential Equations

In differential equations, a solution is a function that satisfies the equation when substituted for the dependent variable. Consider the differential equation given by \((y'')^2 = y^2\). To find solutions, we need functions \(y_1\) and \(y_2\) that, when differentiated and plugged back, hold the equation true. Here, the functions \(y_1 = e^x\) and \(y_2 = \cos x\) are proposed solutions.

Let's summarize the task to verify them:

• For \(y_1 = e^x\), we differentiate: the second derivative is \(y_1'' = e^x\). When substituted into the equation, \((e^x)^2 = e^{2x}\) matches \(y_1^2 = e^{2x}\).
• For \(y_2 = \cos x\), the second derivative is \(y_2'' = -\cos x\). Substitute, and we get \((-\cos x)^2 = \cos^2 x\), which equals \(y_2^2 = \cos^2 x\).

The equation holds true for both functions individually, confirming them as valid solutions.

Verification of Solutions

Verification involves confirming that substitution into the differential equation maintains both sides equal. Here's how we verify if a function is a solution:

• Calculate the respective derivatives needed to plug into the differential equation.
• Substitute these derivatives back into the original equation to check equality.

For instance, with our example, verifying \(y_1 = e^x\) required differentiating twice to get \(y_1'' = e^x\), which led to \(e^{2x} = e^{2x}\), illustrating that both sides of the equation are indeed equal.

Similarly, for \(y_2 = \cos x\), substituting \((-\cos x)^2\) compares with \(\cos^2 x\). This step-by-step substitution is crucial because it ensures that the proposed function satisfies the differential equation.

General Solutions

The general solution of a differential equation involves a combination of particular solutions. Often, the general form like \(y = c_1 y_1 + c_2 y_2\) with constants \(c_1, c_2\), encompasses multiple specific solutions. However, just because \(y_1\) and \(y_2\) are solutions individually, their linear combination isn't automatically a solution to the same differential equation.

In our exercise, the general solution \(y = c_1 e^x + c_2 \cos x\) was proposed. Differentiating gives \(y'' = c_1 e^x - c_2 \cos x\). Substituting into \((y'')^2 = y^2\) doesn't yield equality unless special conditions like \(c_1 = 0\) or \(c_2 = 0\) are met.

This teaches us that while general solutions broaden possibilities, they need careful verification against specific equations to confirm their validity. Each problem is unique and might require tailored solutions.