Question

Find the solution of the given initial value problem. Then plot a graph of the solution. \(y^{\prime \prime \prime}+4 y^{\prime}=t, \quad y(0)=y^{\prime}(0)=0, \quad y^{\prime \prime}(0)=1\)

Short answer

Question: Determine the solution to the linear third-order ordinary differential equation $y^{\prime \prime \prime} + 4y^{\prime} = t$ with the initial conditions $y(0)=0$, $y^{\prime}(0)=0$, and $y^{\prime \prime}(0)=1$. Plot the graph of the solution. Answer: The solution to the given differential equation with the initial conditions is $y(x) = -\frac{1}{4} \cos(2x) + \frac{1}{8}x^2$. The graph of this function can be plotted using graphing tools such as Wolfram Alpha, Desmos, or a graphing calculator.

Step-by-step solution

Find the auxiliary equation and its roots

Start by finding the auxiliary equation to the homogeneous part of the given ODE, which is: $$y^{\prime \prime \prime} + 4y^{\prime} = 0$$ The auxiliary equation is: $$m^3 + 4m = 0$$ Factor out a common factor of \(m\), giving: $$m(m^2 + 4) = 0$$ Thus, we get the roots: \(m_1 = 0\), \(m_2 = 2i\), and \(m_3 = -2i\).

Construct the complementary function

Now, construct the complementary function (CF) from these roots, which will be the general solution of the homogeneous equation. The complementary function is: $$\text{CF} = C_1 + C_2 \cos(2x) + C_3 \sin(2x)$$

Find the particular solution

We are now going to find the particular solution of the given inhomogeneous equation: $$y^{\prime \prime \prime} + 4y^{\prime} = t$$ As the right-hand side of the equation is a polynomial, we can assume a particular solution of the form: $$y_p = At^2 + Bt + C$$ Compute the first, second, and third derivatives: $$y_p^{\prime} = 2At + B$$ $$y_p^{\prime \prime} = 2A$$ $$y_p^{\prime \prime \prime} = 0$$ Substitute these derivatives into the given inhomogeneous equation and solve for the constants A, B, and C: $$0 + 4(2At + B) = t$$ $$8At + 4B = t$$ Comparing the coefficients, we get \(A = \frac{1}{8}\) and \(B = 0\). So, the particular solution is: $$y_p = \frac{1}{8}t^2$$

Combine complimentary function and particular solution

Now we have both the complementary function and the particular solution, we can write the general solution as: $$y(x) = C_1 + C_2 \cos(2x) + C_3 \sin(2x) + \frac{1}{8}t^2$$

Apply the initial conditions

We are given the initial conditions \(y(0)=0\), \(y^{\prime}(0)=0\), and \(y^{\prime \prime}(0)=1\). Apply these conditions to the general solution and solve for the constants \(C_1\), \(C_2\), and \(C_3\): $$y(0) = C_1 + \frac{1}{8}(0)^2 = 0 \Rightarrow C_1 = 0$$ $$y^{\prime}(0) = 2\frac{1}{8}(0) + C_2 \cdot (-2\sin(0)) + C_3 \cdot 2\cos(0) = 0 \Rightarrow C_3 = 0$$ $$y^{\prime \prime}(0) = 2\frac{1}{8} - 4C_2 \cos(0) = 1 \Rightarrow C_2 = -\frac{1}{4}$$ So, the final solution with the initial conditions is: $$y(x) = -\frac{1}{4} \cos(2x) + \frac{1}{8}x^2$$

Plot the graph of the solution

Finally, we'll plot the graph of the solution. Take the function $$y(x) = -\frac{1}{4} \cos(2x) + \frac{1}{8}x^2$$ and plot its graph using any graphing tool or software such as Wolfram Alpha, Desmos, or a graphing calculator.

Key concepts

Auxiliary Equation

Learning how to solve differential equations often starts with understanding the auxiliary equation. It's an essential component in dealing with higher-order linear differential equations. Imagine you're given a differential equation that involves finding the solution to its homogeneous part. The homogeneous form ignores any external influence or forcing function (i.e., the non-zero right side of the original equation).

Here's where the auxiliary equation steps in. The process involves assuming a solution of an exponential form, typically of the nature \(y = e^{mx}\), and substituting it into the homogeneous equation. This substitution helps simplify the equation into a polynomial form that can be solved for \(m\), the roots of which will lead us to recognize the nature of the solutions.

In our example, the auxiliary equation for the homogeneous counterpart of the given inhomogeneous differential equation is \(m^3 + 4m = 0\). By factoring and solving it, we gather information about the structure of the complementary function.

Complementary Function

The complementary function, also known as the general solution of the homogeneous equation, is built from the roots of the auxiliary equation. It captures the inherent characteristics of the system defined by the differential equation.

In simpler terms, it tells you how the solution behaves in the absence of external forces or inputs. Depending on the type of roots—real or complex—the structure of the complementary function may vary.

In our case, we had roots \(m_1 = 0\), \(m_2 = 2i\), and \(m_3 = -2i\). Here's how these roots map to the complementary function:

• The real root, \(m_1 = 0\), corresponds to a constant solution, \(C_1\).
• The complex roots, \(2i\) and \(-2i\), suggest oscillatory components expressed by \(C_2 \cos{2x}\) and \(C_3 \sin{2x}\).

Thus, our complementary function becomes \(C_1 + C_2 \cos(2x) + C_3 \sin(2x)\).

Particular Solution

Finding the particular solution is about incorporating the effects of external inputs or forcing functions into our differential equation. It's key to solving inhomogeneous differential equations where there is a non-zero right-hand side, such as a forcing function, external force, or input.

In our problem, the non-zero side is a simple polynomial function, \(t\), which guided our choice in assuming a form for the particular solution. We assumed it in a polynomial format as well: \(y_p = At^2 + Bt + C\).

Obtaining the particular solution required us to differentiate this assumed polynomial form, substitute back into our differential equation, and then equate coefficients to find the constants. This step-by-step process is crucial in tailoring solutions that respect both the inherent system dynamics (complementary function) and the external influences (particular solution).

In the example, we find that \(A = \frac{1}{8}\) and \(B = 0\), resulting in a particular solution \(y_p = \frac{1}{8}t^2\).

Inhomogeneous Differential Equation

An inhomogeneous differential equation is an equation that resembles its homogeneous counterpart, but with an added forcing function or external input expressed on the right-hand side. This additional term shifts the nature of the solutions, as it might represent real-world influences like sinusoidal signals, exponential growth, or polynomial trends.

Inhomogeneous equations are pivotal in modeling situations where systems are externally driven, as in the case of forced oscillations or electric circuits with alternating current. For the given exercise, the external factor is present as \(\sec t\), which requires finding both the complementary function and the particular solution to fully address the problem.

By solving the homogeneous equation for the complementary function and then solving the whole inhomogeneous equation for the particular solution, we merge these insights to form a thorough general solution. This solution beautifully incorporates the natural behavior of the system, and how it reacts to external inputs.