Question

If \(\frac{d^{2} y}{d x^{2}}=x-5\) and at \(x=0, y^{\prime}=-2\) and \(y=1,\) find a solution of the differential equation.

Short answer

Question: Solve the given second-order differential equation and find the solution that satisfies the given initial conditions: \( \frac{d^2y}{dx^2} = x-5 \), \(y(0)=1\), and \(y'(0)=-2\). Answer: The solution of the given second-order differential equation that satisfies the given initial conditions is \(y(x) = \frac{1}{6}x^3 - \frac{5}{2}x^2 -2x + 1\).

Step-by-step solution

Integrate the given second-order differential equation once

To find the first-order differential equation, we'll integrate \(\frac{d^2y}{dx^2} = x-5\) with respect to \(x\). Let \(y' = \frac{dy}{dx}\). Then: $$\int \frac{d^2y}{dx^2} dx = \int (x-5) dx$$ Integrate on both sides: $$y'(x) = \frac{1}{2}x^2 - 5x + C_1$$ Where \(C_1\) is the integration constant.

Integrate the first-order differential equation

Now, we'll integrate the first-order differential equation \(y'(x)= \frac{1}{2}x^2 - 5x + C_1\) with respect to \(x\). $$\int y'(x) dx = \int \left(\frac{1}{2}x^2 - 5x + C_1\right) dx$$ Integrate on both sides: $$y(x) = \frac{1}{6}x^3 - \frac{5}{2}x^2 + C_1x + C_2$$ Where \(C_2\) is another integration constant.

Use the initial conditions to find the values of \(C_1\) and \(C_2\)

We are given that at \(x=0\), \(y(0)=1\) and \(y'(0)=-2\). Let's use these conditions to find the constants \(C_1\) and \(C_2\): For \(x=0\), we have: $$y(0) = \frac{1}{6}(0)^3 - \frac{5}{2}(0)^2 + C_1(0) + C_2 = 1 \Rightarrow C_2 = 1$$ Now for \(y'(0)=-2\): $$y'(0) = \frac{1}{2}(0)^2 - 5(0) + C_1 = -2 \Rightarrow C_1 = -2$$

Write the final solution

Now that we have the values of the constants, we can write the final solution of the differential equation: $$y(x) = \frac{1}{6}x^3 - \frac{5}{2}x^2 -2x + 1$$ This is the solution of the given second-order differential equation that satisfies the given initial conditions.

Key concepts

Differential Equation Integration

When we talk about differential equation integration, we are referring to the process of finding a function that satisfies a differential equation. In the context of a second-order differential equation like \(\frac{d^{2} y}{d x^{2}}=x-5\), the integration process involves finding the antiderivative twice. First, we integrate to get a first-order differential equation.

For our example, the result of the first integration \(\int (x - 5) dx\) gives us \(y'(x) = \frac{1}{2}x^2 - 5x + C_1\), where \(C_1\) is an integration constant. The second integration of \(y'(x)\) leads us to the general solution \(y(x) = \frac{1}{6}x^3 - \frac{5}{2}x^2 + C_1x + C_2\) with another constant \(C_2\). Each integration step reduces the order of the derivative by one.

Initial Conditions in Differential Equations

The concept of initial conditions is crucial for uniquely determining the solution to a differential equation. Initial conditions are specific values of the function and/or its derivatives at a particular point. These conditions allow us to find the exact values of the integration constants we introduced during the integration process.

In our exercise, we are provided with two initial conditions at \(x=0\): \(y'(0) = -2\) and \(y(0) = 1\). Using these conditions, we can solve for \(C_1\) and \(C_2\). Substituting \(x = 0\) into the integrated equations, we obtain \(C_2 = 1\) and \(C_1 = -2\). With the constants determined, we have found a unique solution that specifically satisfies both the differential equation and the initial conditions.

First-Order Differential Equation

A first-order differential equation involves the first derivative of the unknown function but no higher derivatives. The general form can be written as \(\frac{dy}{dx} = f(x)\).

The integration of a first-order differential equation is a straightforward process. For the exercise at hand, the first integration of the given second-order differential equation produced a first-order differential equation \(y'(x) = \frac{1}{2}x^2 - 5x + C_1\). Solving first-order differential equations is a foundational technique needed to address more complex, higher-order cases.

Second-Order Differential Equation

A second-order differential equation involves the second derivative of the function. The general form of such an equation is \(\frac{d^{2}y}{dx^{2}} = g(x)\). They are more complex than first-order differential equations because they require the integration process to be performed twice, each producing an integration constant.

In our case, the starting equation \(\frac{d^{2} y}{d x^{2}}=x-5\) is a second-order linear differential equation with constant coefficients. The solution process as demonstrated in the steps above, results in the final solution \(y(x) = \frac{1}{6}x^3 - \frac{5}{2}x^2 -2x + 1\) after integrating twice and applying the initial conditions.