Question

Find the solution of the given initial value problem. Then plot a graph of the solution. $y^{\mathrm{iv}}+2y^{\prime \prime }+y=3t+4,y(0)=y^{\prime }(0)=0,y^{\prime \prime }(0)=y^{\prime \prime \prime }(0)=1$

Short answer

Based on the steps provided in the solution, summarize your understanding of the process followed to find the unique solution of the given DE. First, the characteristic equation was identified and its roots were found. Next, the complementary function and particular solution were written down. The complementary function and the particular solution were then combined to find the general solution of the DE. After that, the given initial conditions were used to find the unique solution. Finally, the unique solution was plotted using graphing software. Overall, the unique solution was found to be: $$y(t)=e^{-t}+t{e}^{-t}+\frac{1}{5}\sin t+\frac{2}{5}\cos t$$

Step-by-step solution

Identify the characteristic equation

The given DE is: $$y^{\mathrm{w}}+2y^{\prime \prime }+y=\sin t$$ The characteristic equation for this DE is: $$r^3+2r^2+r=0$$

Find the roots of the characteristic equation

First, factor out r from the equation: $$r(r^2+2r+1)=0$$ Now we have a quadratic equation inside the parenthesis: $$r^2+2r+1=0$$ This is a perfect square trinomial, which can be expressed as: $$(r+1{)}^2=0$$ So, the roots are r=-1 (triple root).

Write the complementary function

Since we have a triple root, the complementary function (CF) will be in the form: $$CF(y)=C_1{e}^{-t}+C_{2}t{e}^{-t}+C_3{t}^2{e}^{-t}$$

Write down the particular solution

Since the DE involves a sinusoidal function, we'll consider the particular solution (PS) in the form: $$PS(y)=A\sin t+B\cos t$$ Now we need to find A and B. Differentiate PS(y) twice with respect to t: $$\begin{gathered} P{S}^{\prime }(y)=A\cos t-B\sin t \\ P{S}^{\prime \prime }(y)=-A\sin t-B\cos t \end{gathered}$$ Now, substitute the PS equation into the given DE. $$(-A\sin t-B\cos t)+2(-A\sin t-B\cos t)+(A\sin t+B\cos t)=\sin t$$ Combine the sine and cosine terms: $$(-A+2B)\sin (t)+(-B-2A)\cos (t)=\sin t$$ Comparing coefficients: $$\begin{gathered} -A+2B=1, \\ -B-2A=0. \end{gathered}$$ Solve the system of linear equations: $$\begin{gathered} A=\frac{1}{5} \\ B=\frac{2}{5} \end{gathered}$$ Now, the particular solution is: $$PS(y)=\frac{1}{5}\sin t+\frac{2}{5}\cos t$$

Combine the complementary function and the particular solution

The general solution of the DE given the initial conditions is: $$y(t)=CF(y)+PS(y)=C_1{e}^{-t}+C_{2}t{e}^{-t}+C_3{t}^2{e}^{-t}+\frac{1}{5}\sin t+\frac{2}{5}\cos t$$

Use the given initial conditions to find the unique solution

The initial conditions are: $$\begin{gathered} y(0)=2, \\ {y}^{\prime }(0)=0, \\ {y}^{\prime \prime }(0)=-1, \\ {y}^{\prime \prime \prime }(0)=1 \end{gathered}$$ Applying the initial conditions and solving for the constants: $$\begin{gathered} C_1+\frac{1}{5}\sin (0)+\frac{2}{5}\cos (0)=2 \\ {C}_2-C_1+\frac{1}{5}\cos (0)-\frac{2}{5}\sin (0)=0 \\ 2C_2-2C_1+\frac{1}{5}\sin (0)-\frac{2}{5}\cos (0)=-1 \\ -C_2-2C_3-\frac{1}{5}\cos (0)-\frac{2}{5}\sin (0)=1 \end{gathered}$$ Solving this system of linear equations, we find: $$\begin{gathered} C_1=1, \\ {C}_2=1, \\ {C}_3=0. \end{gathered}$$ Now the unique solution is $$y(t)=e^{-t}+t{e}^{-t}+\frac{1}{5}\sin t+\frac{2}{5}\cos t$$

Plot a graph of the solution

To plot a graph of the solution, we can use graphing software (e.g., Desmos, Wolfram Alpha, etc.) with the equation: $$y(t)=e^{-t}+t{e}^{-t}+\frac{1}{5}\sin t+\frac{2}{5}\cos t$$ A graph of the solution can be obtained through any online plotting tools by inputting the given function.

Key concepts

Differential Equations

Differential equations are mathematical equations that involve derivatives of a function. In simpler terms, they express how a quantity changes with respect to others. These equations are crucial because they can model real-world situations like the growth of populations, the motion of waves, or even electrical circuits. For example, our given problem involves the differential equation:$$y^{iv}+2y^{″}+y=3t+4$$Here, the function $y(t)$ depends on the variable $t$, and the equation involves derivatives such as $y^{″}$ (the second derivative of $y$) and $y^{iv}$ (the fourth derivative of $y$). Solving a differential equation means finding the function $y(t)$ that satisfies this relationship for every $t$. Solutions can be unique or general, depending on initial or boundary conditions provided. Initial conditions, such as $y(0)=y^{\prime }(0)=0$, help us find a particular solution that fits our specific scenario.

Particular Solution

A particular solution is a specific solution to a differential equation that satisfies given conditions, such as initial values or boundary conditions. In the context of our exercise, we aim to find a function $y(t)$ that not only satisfies the equation $y^{iv}+2y^{″}+y=3t+4$ but also meets the conditions $y(0)=y^{\prime }(0)=0$ and $y^{″}(0)=y^{‴}(0)=1$.To find the particular solution, we need to consider the non-homogeneous part of the differential equation, in this case, the function $3t+4$. A technique called "undetermined coefficients" or "variation of parameters" can be used, where we assume a form of $y(t)$ and find coefficients that satisfy the differential equation. Once these coefficients are determined, we combine the particular solution with the complementary function (discussed in the next section) to form the general solution, and then apply the initial conditions to find the specific particular solution that satisfies the problem entirely.

Complementary Function

The complementary function (CF) is an essential component of solving linear differential equations. It represents the solution to the corresponding homogeneous differential equation, where the right-hand side of the equation is zero. In other words, it's like finding the natural response of the system without any external influence.For our problem, the homogeneous differential equation is:$$y^{iv}+2y^{″}+y=0$$To solve it, we start by finding the characteristic equation, which involves replacing each derivative with powers of a variable, say $r$. This translates to:$$r^4+2r^2+r=0$$By solving this equation, we find roots (values of $r$) that help construct the CF:$$CF(y)=C_1{e}^{-t}+C_{2}t{e}^{-t}+C_3{t}^2{e}^{-t}$$where $C_1$, $C_2$, and $C_3$ are arbitrary constants. This formula comprises terms that fade away over time (represented by $e^{-t}$) and may be multiplied by polynomial terms depending on the multiplicity of the roots, indicating repeated solutions.

Characteristic Equation

A characteristic equation is a tool to find the roots of the differential equation, particularly for its complementary function. It involves transforming the differential equation into a polynomial equation using a substitution method. Each derivative in the differential equation is replaced with powers of a variable $r$, creating a polynomial equation that is easier to handle.For the differential equation given:$$y^{iv}+2y^{″}+y=0$$The characteristic equation becomes:$$r^4+2r^2+r=0$$By factoring this polynomial, we can locate its roots. In this case, the factorization results in roots such as $r=-1$ with multiplicity, leading to a characteristic equation as a set of linear factors that directly influence the form of the complementary function. Solving this algebraic equation is key to identifying the equation's fundamental solutions and, ultimately, forming the complementary function needed to obtain the general solution of the differential equation.