Question

Second-Order Differential Equations Find the general so- lution to each of the following second-order differential equa- tions by first finding \(d y / d x\) and then finding \(y\) . The general solu- tion will have two unknown constants. (a) \(\frac{d^{2} y}{d x^{2}}=12 x+4\) (b)\(\frac{d^{2} y}{d x^{2}}=e^{x}+\sin x\) (c) \(\frac{d^{2} y}{d x^{2}}=x^{3}+x^{-3}\)

Short answer

The general solutions to the differential equations are (a) \[y=2x^{3}+2x^{2}+C1 * x+C2\], (b) \[y=e^{x}-\sin x+C1 * x + C2\], and (c) \[y= \frac{1}{20}x^{5}+1/4x^{-1}+C1 * x + C2\]. \(C1\) and \(C2\) are constants which are determined by initial conditions.

Step-by-step solution

Finding \(d y / d x\) of equation (a)

The given equation is \(\frac{d^{2} y}{d x^{2}}=12 x+4\). Use indefinite integration to find \(d y / d x\). This results in \(d y / d x= 6x^{2} +4x+ C1\), where \(C1\) is a constant because when differentiating a constant, the result is zero.

Finding \(y\) of equation (a)

Once more use indefinite integration to find \(y\). This results in \(y = 2x^{3}+2x^{2} + C1 * x + C2\), where \(C1\) and \(C2\) are constants.

Finding \(d y / d x\) of equation (b)

For the equation \(\frac{d^{2} y}{d x^{2}}=e^{x}+\sin x\), use indefinite integration. This results in \(d y / d x= e^{x}-\cos x+ C1\), where \(C1\) is a constant.

Finding \(y\) of equation (b)

Use indefinite integration to find \(y\). This results in \(y = e^{x}-\sin x + C1 * x + C2\), where \(C1\) and \(C2\) are constants.

Finding \(d y / d x\) of equation (c)

The given equation is \(\frac{d^{2} y}{d x^{2}}=x^{3}+x^{-3}\). Use indefinite integration to find \(d y / d x\). This results in \(d y / d x= \frac{1}{4}x^{4}-\frac{1}{2}x^{-2}+ C1\), where \(C1\) is a constant.

Finding \(y\) of equation (c)

Once more use indefinite integration to find \(y\). This results in \(y = \frac{1}{20}x^{5}+1/4x^{-1} + C1 * x + C2\), where \(C1\) and \(C2\) are constants.

Key concepts

Indefinite Integration

Indefinite integration, often referred to simply as integration, is a core operation in calculus used to construct antiderivatives. An antiderivative of a function is another function that, when differentiated, gives the original function. When solving second-order differential equations, like the ones presented in the exercise, indefinite integration is used to find the first derivative \( d y / d x \) from the second derivative \( d^{2} y / d x^{2} \).

This process involves finding a function whose derivative matches the given function. For example, to integrate the function \(12x+4\), one might recall that the derivative of \(x^n\) is \(nx^{n-1}\) and work backwards, resulting in an antiderivative of \(6x^2+4x\). However, since differentiation of a constant yields zero, the integration process introduces an arbitrary constant, often denoted as \(C\). This is because an indefinite integral indicates a family of functions, all differing by a constant.

General Solution of Differential Equations

The general solution to a differential equation contains all possible solutions to the equation. For instance, when solving second-order differential equations like the ones given in the exercise, the general solution will incorporate two indefinite integrations, each introducing a constant (\(C1\) and \(C2\)).

After integrating the given second derivative twice, we obtain an expression for \(y\) that includes two integration constants. These constants represent the infinite family of possible solutions to the equation and are determined when specific initial conditions or boundary conditions are provided for a particular problem. Without such conditions, the solution remains general—an equation that includes conditions for all possible specific solutions.

Integration Constants

Integration constants are of paramount importance in the process of indefinite integration. They appear as a result of the \( + C \) term, which signifies an unknown constant. Each act of indefinite integration introduces another such constant into the solution of a differential equation.

When solving a second-order differential equation, we integrate twice and, as a result, we get two constants (\(C1\) and \(C2\)). These constants are crucial for the solution to accurately describe all possible antiderivatives of the function, and they are ultimately what make the solution a 'general' one. Whenever initial value conditions are known, they can be substituted into the general solution to solve for these constants, thus whittling down the infinite set of solutions to the one specific solution that fits the given scenario.

Differential Equation Integration Strategies

When integrating to solve differential equations, different strategies are employed depending on the form of the equation. In the provided exercise, we can see varied examples. For the linear polynomial \(12x+4\), a simple power rule is employed where one adds one to the exponent then divides by the new exponent. With exponential and trigonometric functions, like \(e^x + sin x\), one uses the basic antiderivatives of these functions, knowing that the integral of \(e^x\) is \(e^x\), and the integral of \(sin x\) is \( -cos x\). And for the algebraic function with positive and negative powers \(x^3+x^{-3}\), each term is dealt with separately according to power rule.

The selection of the right integration technique is crucial and may include methods such as substitution, integration by parts, or even advanced strategies for more complex functions. These strategies enable one to tackle a wide range of differential equations, but each particular equation may present its own challenges, requiring a bespoke approach.