Question

Find the solution of each of the following initial-value problems: (a) \(\quad y^{\prime \prime}+4 y^{\prime}+13 y=0\), \(y(0)=0, \quad y^{\prime}(0)=-1 .\) (b) \(\quad y^{\prime \prime}+y=e^{-2 t} \sin t\) \(y(0)=0, \quad y^{\prime}(0)=0 .\)

Short answer

Question: Solve the given initial-value problems: (a) \(y^{\prime\prime}+4y^{\prime}+13y=0\), \(y(0)=0\), \(y^{\prime}(0)=-1\) (b) \(y^{\prime\prime}+y=e^{-2t}\sin t\), \(y(0)=0\), \(y^{\prime}(0)=0\) Answer: (a) \(y(t)=-\frac{1}{3}e^{-2t}\sin(3t)\) (b) \(y(t)=\frac{1}{\sqrt{5}}\sin t-\frac{1}{\sqrt{5}}te^{-2t}\sin t\)

Step-by-step solution

Find the complementary function for the homogeneous equation

Write down the given homogeneous linear differential equation and find its characteristic equation. The given differential equation is \(y^{\prime\prime}+4y^{\prime}+13y=0\). The characteristic equation will be: \(m^2+4m+13=0\).

Solve the characteristic equation

Use the quadratic formula to find the roots of the characteristic equation. For a quadratic equation \(ax^2+bx+c=0\), the roots are given by \(\displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\). Plugging the values into the quadratic formula, we get: \(m_{1,2}=\displaystyle\frac{-4\pm\sqrt{4^2-4\cdot1\cdot13}}{2\cdot1}=\frac{-4\pm\sqrt{-36}}{2}=-2\pm3i\).

Write down the general solution for the homogeneous equation

Based on the complex roots, write the general solution for the homogeneous linear differential equation. Since we have complex conjugate roots, the general solution is of the form: \(y(t)=e^{-2t}(c_1\cos(3t)+c_2\sin(3t))\).

Apply the initial conditions

Apply the initial conditions to find the particular solution. We are given \(y(0)=0\) and \(y^{\prime}(0)=-1\). First, plug in \(t=0\) into the general solution to find \(c_1\). \(0=e^{-2\cdot0}(c_1\cos(3\cdot0)+c_2\sin(3\cdot0))=e^0(c_1\cos(0)+c_2\sin(0))=c_1\implies c_1=0\). Now, plug in \(t=0\) into the derivative of the general solution, since \(y^{\prime}(0)=-1\). First, find the derivative: \(y^{\prime}(t)=-2e^{-2t}(c_1\cos(3t)+c_2\sin(3t))+e^{-2t}\cdot(-3c_1\sin(3t)+3c_2\cos(3t))\) Plug in \(t=0\): \(-1=-2e^0(c_1\cos(0)+c_2\sin(0))+e^0\cdot(-3c_1\sin(0)+3c_2\cos(0))=-2c_1+3c_2\implies c_2=-\frac{1}{3}\).

Write down the particular solution

Plug in the values of \(c_1\) and \(c_2\) into the general solution and simplify to get the final answer. \(y(t)=e^{-2t}(0\cos(3t)-\frac{1}{3}\sin(3t))=-\frac{1}{3}e^{-2t}\sin(3t)\). Thus, the solution for part (a) is: \(\displaystyle y(t)=-\frac{1}{3}e^{-2t}\sin(3t)\). ##Part (b)##

Find the complementary function for the homogeneous equation

Write down the given homogeneous linear differential equation and find its characteristic equation. The given non-homogeneous differential equation is \(y^{\prime\prime}+y=e^{-2t}\sin t\). The corresponding homogeneous equation is \(y^{\prime\prime}+y=0\). The characteristic equation will be: \(m^2+1=0\).

Solve the characteristic equation

Find the roots of the characteristic equation. The roots are \(m_{1,2}=\pm i\).

Write down the complementary function

Based on the complex roots, write down the complementary function for the homogeneous linear differential equation. Since the roots are complex and \(y_p(t)\neq c(t)e^{-2t}\sin t \), the complementary function is: \(y_c(t)=c_1\cos t+c_2\sin t\).

Find the particular solution using the method of undetermined coefficients

Assume \(y_p(t)=t(At+B)e^{-2t}\sin t\). Now find \(y_p^{\prime}(t)\) and \(y_p^{\prime\prime}(t)\), and plug these into the given non-homogeneous differential equation. We have \(y_p(t)=t(At+B)e^{-2t}\sin t\), so \(y_p'(t)=e^{-2t}\sin t[(-2t)(At+B)+t(A)]+t(At+B)e^{-2t}\cos t\) and \(y_p''(t)=e^{-2t}\sin t[4t(At+B)-6(A+B)t+A]+2e^{-2t}\cos t[-2t(At+B)+A]-t(At+B)e^{-2t}\sin t\). Now plug \(y_p^{\prime}(t)\) and \(y_p^{\prime\prime}(t)\) into the non-homogeneous equation: \(y^{\prime\prime}+y=e^{-2t}\sin t\Longrightarrow [4t(At+B)-6(A+B)t+A]+2(-2t(At+B)+A\cos t)-t(At+B)(\sin t)=(e^{-2t}\sin t)\). By comparing the coefficients of the sine and cosine terms, we obtain the following system of equations: Left-side coefficients of \(\sin t\): \(4t^2A-6At+A-t^2A=t^2A-6At+A=1\) Left-side coefficients of \(\cos t\): \(2(-2tB+2A)=4A-4tB=0\)

Solve the system of equations to find the coefficients

Solve the system of equations to find the values of A and B. Solve the second equation for A: \(A=tB\). Now, plug this into the first equation: \(t^2(tB)-6A(tB)+tB=1\rightarrow t^3B^2-6t^2B^2+tB=1\). Since \(t\neq 0\), we divide by t and simplify for B: \(t^2B^2-6tB^2+B=1\rightarrow B^2(t^2-6t)=1\rightarrow B=\pm\sqrt{\frac{1}{t^2-6t}}\). Since \(B\) is a constant, we choose the non-zero root for \(t\): \(B=-\frac{1}{\sqrt{5}}\). And then \(A=-\frac{t}{\sqrt{5}}\).

Write down the particular solution

Use the coefficients A and B to find the particular solution for the non-homogeneous linear differential equation. \(y_p(t)=t(-\frac{t}{\sqrt{5}}+\frac{1}{\sqrt{5}})e^{-2t}\sin t = -\frac{1}{\sqrt{5}}te^{-2t}\sin t\).

Write the general solution

Combine the complementary function and the particular solution to obtain the general solution for the non-homogeneous equation. \(y(t)=y_c(t)+y_p(t)=c_1\cos t+c_2\sin t-\frac{1}{\sqrt{5}}te^{-2t}\sin t\).

Apply the initial conditions

Apply the initial conditions to find the particular solution. We are given \(y(0)=0\) and \(y^{\prime}(0)=0\). First, let's find \(y^{\prime}(t)=(-c_1\sin t+c_2\cos t)-\frac{1}{\sqrt{5}}[(2t-1)e^{-2t}\sin t+te^{-2t}\cos t]\). Now, plug in the initial conditions: \(y(0)=c_1\cos(0)+c_2\sin(0)-\frac{1}{\sqrt{5}}\cdot0=c_1\); and \(y^{\prime}(0)=-c_1\sin(0)+c_2\cos(0)-\frac{1}{\sqrt{5}}[(2(0)-1)e^{-2\cdot0}\sin(0)+0\cdot e^{-2\cdot0}\cos(0)]=0\cdot c_1+c_2\cdot1-\frac{1}{\sqrt{5}}=c_2\). Thus, \(c_1=0\) and \(c_2=\frac{1}{\sqrt{5}}\).

Write down the particular solution

Plug in the values of \(c_1\) and \(c_2\) into the general solution and simplify to get the final answer. \(y(t)=0\cos t+\frac{1}{\sqrt{5}}\sin t-\frac{1}{\sqrt{5}}te^{-2t}\sin t=\frac{1}{\sqrt{5}}\sin t-\frac{1}{\sqrt{5}}te^{-2t}\sin t\). Thus, the solution for part (b) is: \(\displaystyle y(t)=\frac{1}{\sqrt{5}}\sin t-\frac{1}{\sqrt{5}}te^{-2t}\sin t\).

Key concepts

Initial-Value Problems

Initial-value problems involve finding a specific solution to a differential equation that satisfies given conditions at a particular point, often

• These conditions specify values for the function and potentially its derivatives at a designated point, generally at the beginning or 'initial' point.
• They are of crucial importance in finding unique solutions, thermodynamics, population modeling, and more.

For example, in the exercise, we had initial conditions like \(y(0)=0\) and \(y'(0)=-1\). Applying these conditions constrains the possible solutions, leading to one specific solution.

Homogeneous Differential Equations

A homogeneous differential equation is a type of differential equation where all terms are dependent solely on the function and its derivatives, without an independent term. In mathematical terms:

• These equations can be expressed as \(L(y)=0\), where \(L\) is a linear differential operator on \(y\).
• In the problem's case, the equation \(y^{\prime \prime} + 4y^{\prime} + 13y = 0\) is homogeneous because it has no independent term.

Such equations are well-analyzed using the characteristic equation, which provides insight into the nature of the solution by focusing on the roots of the equation.

Non-Homogeneous Differential Equations

Non-Homogeneous Differential Equations include an independent term, resulting in an equation having the form \(L(y) = g(t)\), where \(g(t)\) isn't uniform.

• One often addresses these using an approach involving both a complementary function for the homogeneous part and a particular solution accounting for the non-homogeneous term.
• In the given problem, the equation \(y^{\prime \prime} + y = e^{-2t} \sin t\) is non-homogeneous due to the presence of \(e^{-2t} \sin t\).

Solutions to such equations typically require techniques like the method of undetermined coefficients to find a solution that satisfies the entire equation.

Complex Roots

Complex roots in the context of differential equations arise when solving the characteristic equation of a linear differential equation, yielding roots of the form \(a \pm bi\).

• These roots lead to oscillatory solutions, typically comprising exponential terms multiplied by trigonometric components.
• In our exercise, the characteristic equation solution produced roots \(-2 \pm 3i\), indicating the solution involves functions like \(e^{-2t}\cos(3t)\) and \(e^{-2t}\sin(3t)\).

Complex roots vividly impact the nature of solutions, especially when dealing with phenomena involving periodic behavior or wave processes.

Characteristic Equation

A characteristic equation is a polynomial equation derived from a linear differential equation with constant coefficients, essential for studying its solutions.

• This equation is obtained by assuming solutions of specific exponential forms and solving for characteristic roots.
• For example, in the problem, the characteristic equation \(m^2 + 4m + 13 = 0\) led us to identify the roots necessary for the homogeneous solution's form.

By understanding these roots' nature, one predicts the fundamental type (real, complex, repeated) of the solution, aiding in forming the general solution.

Method of Undetermined Coefficients

The Method of Undetermined Coefficients is a strategy to find particular solutions to non-homogeneous linear differential equations with constant coefficients.

• It works by positing a trial solution that resembles the non-homogeneous term's form \(g(t)\), and then identifying the necessary coefficients.
• In the problem, to solve \(y^{\prime \prime} + y = e^{-2t} \sin t\), a particular part \(y_p\) form \(t(At + B)e^{-2t} \sin t\) was used.

This method simplifies solving complex differential equations by transforming the process into one of solving a system of algebraic equations for coefficients.