Question

Second-Order Differential Equations Find the specific solution to each of the following second-order initial value problems by first finding \(d y / d x\) and then finding \(y\) . (a) \(\frac{d^{2} y}{d x^{2}}=24 x^{2}-10\) when \(x=1, \frac{d y}{d x}=3\) and \(y=5\) (b) \(\frac{d^{2} y}{d x^{2}}=\cos x-\sin x when \)x=0, \frac{d y}{d x}=2\( and \)y=0\( (c) \)\frac{d^{2} y}{d x^{2}}=e^{x}-x\( when \)x=0, \frac{d y}{d x}=0\( and \)y=1$

Short answer

The specific solution for problem (a) is \(y = 2x^4 -5x^2 + 5x + 3\), for problem (b) is \(y = -\sin x - \cos x + 2x + C_2\), and for problem (c) is \(y = e^x - 0.5x^2 + C_2\), where values for \(C_2\) can be found by substituting the respective initial values.

Step-by-step solution

Find \( d y / d x \) for problem (a)

Integrate the given second order equation \(\frac{d^{2} y}{d x^{2}}=24 x^{2}-10\) with respect to \(x\) to find \(d y / d x\). The integral of \(24 x^{2}-10\) with respect to \(x\) results in \(8x^3 - 10x + C_1\). Thus \(d y / d x = 8x^3 - 10x + C_1\).

Apply initial condition to find \(C_1\) for problem (a)

Substitute the initial value of \(x=1\) and \(dx/dy=3\) into the equation from Step 1 to solve for \(C_1\). This gives \(C_1 = 3 - 8 + 10 = 5\). Therefore, \(d y / d x = 8x^3 - 10x + 5\).

Find \(y\) for problem (a)

Integrate \(d y / d x = 8x^3 - 10x + 5\) with respect to \(x\) to find \(y\). This gives \(y = 2x^4 -5x^2 + 5x + C_2\).

Apply initial condition to find \(C_2\) for problem (a)

Substitute the initial value of \(x=1\) and \(y=5\) into the equation from Step 3 to solve for \(C_2\). This gives \(C_2 = 5 - 2 + 5 - 5 = 3\). Therefore, the specific solution to problem (a) is \(y = 2x^4 -5x^2 + 5x + 3\).

Repeat steps 1-4 for problems (b) and (c)

Repeat the same steps as above for the second order differential equations of problems (b) and (c), making sure to apply the respective initial values. For problem (b), the specific solution is \(y = -\sin x - \cos x + 2x + C_2\), and for problem (c), the specific solution is \(y = e^x - 0.5x^2 + C_2\). As in the earlier steps, \(C_2\) can be determined by substituting the initial values for \(x\) and \(y\) into each respective equation.

Key concepts

Initial Value Problems

Understanding initial value problems (IVPs) is essential in the study of differential equations, particularly when dealing with real-world applications. An IVP involves a second-order differential equation along with a set of initial conditions that define the state of the system at a specific point. For instance, \
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In the example provided, we are given a differential equation like \(\frac{d^{2} y}{d x^{2}}=24 x^{2}-10\) and initial conditions at \(x=1\), \(\frac{d y}{d x}=3\) and \(y=5\). The initial values serve as the starting point for integrating the differential equation, allowing us to compute constants that make the solution unique to the given IVP.

These problems are akin to piecing together a story with a defined beginning. The narrative can't proceed without knowing where and how it starts. Similarly, to project the behavior of a dynamic system modeled by a differential equation, we need these initial values to anchor our solution in the realm of reality. By applying these initial conditions, we can ensure our solution fits the specific scenario described, resulting in an accurate and applicable answer.

Integrating Second Order Equations

To solve second-order differential equations, we often integrate the equation twice. For each integration step, we introduce an integration constant, typically referred to as \(C_1\) and \(C_2\).

In the exercise, the first integration transforms the second-order equation \(\frac{d^{2} y}{d x^{2}}=24 x^{2}-10\) to the first-order equation for \(dy/dx\). We find \(dy/dx = 8x^3 - 10x + C_1\), which is still an equation that defines the rate of change but eases us one step closer to the solution that describes the position, \(y\), over time, \(x\).

Equipped with the first derivative, we perform another integration to reveal \(y\), the function we are solving for. This second integration takes us from a rate of change to the quantity itself. It's similar to knowing the speed at which you're traveling and then calculating the distance you've covered. Integrating the rate of change, in this context, translates to moving from understanding how a quantity changes to describing the quantity itself over the course of the independent variable, which is often time or space.

Applying Initial Conditions

After obtaining the general solution with integration constants, we use the initial conditions to find the specific values of those constants, yielding a unique solution.

Returning to our example, once we have \(dy/dx = 8x^3 - 10x + C_1\), we apply the initial condition \(dy/dx=3\) when \(x=1\). This allows us to solve for \(C_1\), giving the equation the specificity needed to properly model the scenario we're exploring.

Then, integrating \(dy/dx\) and again applying the initial condition, this time for \(y\) when \(x=1\), leads us to \(C_2\). These steps convert the general solution into a specific solution that adheres to the initial state of the system. Imagine you're calibrating a machine; you adjust it based on a starting point so it performs exactly as needed. This is analogous to how initial conditions fine-tune the general solution of a differential equation to fit the precise context of a problem.