Question

Consider the differential equation \(d^{2} f(x) / d x^{2}=b f(x)\). (a) Suppose that \(f_{1}(x)\) and \(f_{2}(x)\) are solutions. That is, $$ \frac{d^{2} f_{1}(x)}{d x^{2}}=b f_{1}(x) \text { and } \frac{d^{2} f_{2}(x)}{d x^{2}}=b f_{2}(x) $$ Show that the equation also holds when the linear combination \(A_{1} f_{1}(x)+A_{2} f_{2}(x)\) is inserted. (b) Suppose that \(f_{3}(x)\) and \(f_{4}(x)\) are solutions of \(d^{2} f(x) / d x^{2}=b f^{2}(x)\). Is \(A_{3} f_{3}(x)+A_{4} f_{4}(x)\) a solution? Justify your answer.

Short answer

For the given differential equations, we see that the linear combination of the solutions for the first equation is indeed a solution for that equation. This is due to the linearity of the first differential equation. However, for the second differential equation, the linear combination does not satisfy it because the differential equation is not linear.

Step-by-step solution

Prove the Linearity of the First Differential Equation

Insert the given linear combination \(A_1 f_1(x) + A_2 f_2(x)\) into the first differential equation. The second derivative \(d^2/dx^2(A_1 f_1(x) + A_2 f_2(x))\) becomes \(A_1 d^2 f_1(x)/dx^2 + A_2 d^2 f_2(x)/dx^2\) due to linearity of differentiation. By using the given differential equations, it can be substituted by \(A_1 b f_1(x) + A_2 b f_2(x)\), which equals \(b (A_1 f_1(x) + A_2 f_2(x))\), hence confirming that the linear combination of \(f_1(x)\) and \(f_2(x)\) is a solution to the first equation.

Analyze the Second Differential Equation

Now proceed to the second differential equation \(d^2 f(x)/dx^2 = b f^2(x)\) which is inherently nonlinear due to the term \(f^2(x)\). The non-linearity manifests when the function \(f(x)\) is squared, making the linearity properties used in the part (a) no longer applicable.

Test the Linearity of the Second Differential Equation

Attempt to prove that \(A_3 f_3(x) + A_4 f_4(x)\) is a solution in similar way as in part (a). The second derivative \(d^2/dx^2(A_3 f_3(x) + A_4 f_4(x))\) becomes \(A_3 d^2 f_3(x)/dx^2 + A_4 d^2 f_4(x)/dx^2\). But now, substituting with the given differential equations results in \(A_3 b f_3^2(x) + A_4 b f_4^2(x)\), which does not equal \(b (A_3 f_3(x) + A_4 f_4(x))^2\). Hence, the linear combination of the solutions, \(f_3(x)\) and \(f_4(x)\), does not form a solution to the second differential equation.

Key concepts

Linearity of Differential Equations

Understanding the concept of linearity in differential equations is essential for grasping how certain types of problems can be solved. Linearity in this context means that the equation meets two key criteria: it must be homogenous, and it must satisfy the principle of superposition. The latter implies that if you have two solutions to a linear differential equation, any linear combination of these solutions is also a solution.

For instance, consider the differential equation \(d^{2} f(x) / d x^{2}=b f(x)\). If \(f_{1}(x)\) and \(f_{2}(x)\) are solutions, the linearity of differentiation allows us to show that \(A_{1} f_{1}(x) + A_{2} f_{2}(x)\) is also a solution. The process involves taking the second derivative of the linear combination and using the original solutions' derivatives, which should yield a result that is a multiple of the initial linear combination. This characteristic is what makes linear equations more straightforward to handle compared to their nonlinear counterparts.

Second Derivative

The second derivative of a function is a measure of how the rate of change of that function's slope is changing. In simple terms, while the first derivative represents velocity if your function is a position-time graph, the second derivative represents acceleration. This concept is pivotal when dealing with second-order differential equations, where you're solving for a function based on information about its acceleration.

In the exercise, dealing with the second derivative \(d^{2} f(x) / d x^{2}\) allows us to understand how the function \(f(x)\) behaves and changes. For linear differential equations, taking the second derivative of a combination of solutions and then reapplying the equation's criteria confirms the superposition principle. Applying this to the example question illustrates how different constants, \(A_1\) and \(A_2\), scale the respective solutions \(f_1(x)\) and \(f_2(x)\), but the underlying behavior governed by the second derivative remains consistent.

Nonlinear Equations

Nonlinear equations introduce complexity beyond what's found in linear equations due to their non-additive nature. A nonlinear equation involves terms that are not simply the first power of the function or its derivatives, such as equations with terms that product or square the function or its derivatives.

These equations do not adhere to the principle of superposition, meaning a linear combination of solutions isn't guaranteed to be a solution. The exercise illustrated this by showing how \(f_{3}(x)\) and \(f_{4}(x)\), both solutions to a nonlinear equation, cannot be linearly combined to produce a new solution. This is clear once you recognize that \(A_{3} f_{3}^2(x) + A_{4} f_{4}^2(x)\) does not equate to \(b (A_{3} f_{3}(x) + A_{4} f_{4}(x))^2\). Understanding the inherent complexities of nonlinear equations is crucial, as many real-world phenomena are governed by such equations, making their study a significant part of higher mathematics and many fields of science.