Question

Find the function $F$ that satisfies the following differential equations and initial conditions. $$F^{\prime \prime }(x)=1,F^{\prime }(0)=3,F(0)=4$$

Short answer

Question: Determine the specific function F(x) that satisfies the given second-order differential equation $F^{\prime \prime }(x)=1$, and the initial conditions $F^{\prime }(0)=3$ and $F(0)=4$. Answer: The specific function F(x) that satisfies the given differential equation and initial conditions is $F(x)=\frac{1}{2}{x}^2+3x+4$.

Step-by-step solution

Integrate the second-order differential equation once.

In order to obtain the general solution of the given second-order differential equation, $F^{\prime \prime }(x)=1$, we need to integrate both sides with respect to $x$. The indefinite integral of $1$ is simply $x$. So after integrating once, we have: $$F^{\prime }(x)=x+C_1$$

Integrate the first-order result once more.

Now we have a first-order differential equation, and we need to integrate it once more to obtain the function $F(x)$. So, integrate both sides of $F^{\prime }(x)=x+C_1$ with respect to $x$,then we have: $$F(x)=\frac{1}{2}{x}^2+C_{1}x+C_2$$ Now we have the general solution of the given second-order differential equation.

Apply the initial conditions to find the specific solution.

We are given the initial conditions $F^{\prime }(0)=3$ and $F(0)=4$. Let's use these initial conditions to determine the values of $C_1$ and $C_2$. First, plug $x=0$ and $F^{\prime }(0)=3$ into the expression of $F^{\prime }(x)$: $$3=F^{\prime }(0)=0+C_1$$ So, we get $C_1=3$. Next, plug $x=0$ and $F(0)=4$ into the expression of $F(x)$: $$4=F(0)=0+3\cdot 0+C_2$$ So, we get $C_2=4$.

Write down the specific solution.

Now that we have found the values for $C_1$ and $C_2$, we can plug them back into the general solution of the second-order differential equation to obtain the specific solution. So, $$F(x)=\frac{1}{2}{x}^2+3x+4$$. Hence, the function $F(x)$ that satisfies the given differential equation and initial conditions is: $$F(x)=\frac{1}{2}{x}^2+3x+4$$

Key concepts

Second-Order Differential Equation

A second-order differential equation involves derivatives of a function up to the second order. This means it includes the second derivative, denoted as $F^{\prime \prime }(x)$. Such equations often describe physical systems, like oscillations or waves. For instance, the equation $F^{\prime \prime }(x)=1$, means that the rate of change of the rate of change (or acceleration) of the function $F(x)$ is constant.To solve a second-order differential equation, one typically integrates twice to find the original function. This process involves:

• First finding a function for the first derivative ($F^{\prime }(x)$). This is done by integrating the second derivative once.
• Then, integrating again to find the original function $F(x)$.

While integrating, we introduce constants (such as $C_1$ and $C_2$) because indefinite integration includes an arbitrary constant. These constants will later be determined by applying initial conditions.

Initial Conditions

Initial conditions are specific values given for a function and its derivatives at a particular point. They are used to find the particular solution for a differential equation from the general solution. In our exercise, the initial conditions provided are $F^{\prime }(0)=3$ and $F(0)=4$.Let's understand how to apply them:

• The first initial condition, $F^{\prime }(0)=3$, tells us that when $x=0$, the slope of the tangent to the curve $F(x)$ is $3$. This helps in determining the first constant, $C_1$, after the first integration.
• The second initial condition, $F(0)=4$, tells us the value of the function $F$ itself when $x=0$ is $4$. This is used to find the second constant, $C_2$.

Using these conditions, we substitute back into the general solution to find specific values for the constants, allowing us to solve for the particular function that satisfies both the differential equation and the initial conditions.

Integration

Integration is a fundamental technique used to solve differential equations. It involves finding a function whose derivative is the given function or expression. In the context of solving a second-order differential equation, integration is performed twice to retrieve the original function.Here’s how it works step-by-step:

• **First Integration:** Given $F^{\prime \prime }(x)=1$, we integrate to find $F^{\prime }(x)$. This results in $F^{\prime }(x)=x+C_1$. The constant $C_1$ arises from the indefinite integral.


• **Second Integration:** We next integrate $F^{\prime }(x)=x+C_1$ with respect to $x$ to discover $F(x)$. The outcome is $F(x)=\frac{1}{2}{x}^2+C_{1}x+C_2$, introducing another constant $C_2$.

Thus, integration transforms a derivative into a function, allowing us to work back from the rate of change to the original function. By involving constants and initial conditions, we can pinpoint an exact solution tailored to the scenario described by the differential equation.