Chapter 5: Problem 1

Solve the following boundary value problems directly, when possible. a. \(x^{\prime \prime}+x=2, \quad x(0)=0, \quad x^{\prime}(1)=0 .\) b. \(y^{\prime \prime}+2 y^{\prime}-8 y=0, \quad y(0)=1, \quad y(1)=0\). c. \(y^{\prime \prime}+y=0, \quad y(0)=1, \quad y(\pi)=0\).

Short Answer

Expert verified

The general solutions for the given differential equations are: (a) \(x(t)=-2cos(t)+2sin(t)+2\), (b) \(y(t)=\frac{1}{3}e^{2t}+\frac{2}{3}e^{-4t}\), and (c) \(y(t)=cos(t)\)

Step by step solution

Solve Equation (a)

The equation \(x^{\prime \prime}+x=2\) is a non-homogeneous differential equation. The general solution is formed of a homogeneous solution and a particular solution. From characteristic equation \(m^2+1=0\), we get \(m=\pm i\). Hence, the homogeneous solution is \(x_h(t)=C_1cos(t)+C_2sin(t)\). The particular solution can be guessed as \(x_p=a\). Subsituting \(x_p\) into the non-homogeneous differential equation yields \(a=2\). So, the general solution is \(x(t)=C_1cos(t)+C_2sin(t)+2\). Apply the initial conditions \(x(0)=0\) gives \(C_1= -2\). Also, \(x^{\prime}(1)=0\), gives \(C_2=2\). Therefore, the solution is \(x(t)=-2cos(t)+2sin(t)+2\)

Solve Equation (b)

The equation \(y^{\prime \prime}+2 y^{\prime}-8 y=0\) is a homogeneous differential equation. From the characteristic equation \(m^2+2m-8=0\), we find \(m_1=2, m_2=-4\). Hence, the general solution is \(y(t)=C_1e^{2t}+C_2e^{-4t}\). Applying the boundary conditions gives us \(C_1+C_2=1\) for \(y(0)=1\) and \(C_1e^{2}+C_2e^{-4}=0\) for \(y(1)=0\). Solving these simultaneous equations yields \(C_1=1/3\) and \(C_2=2/3\). Therefore, the solution to the differential equation is \(y(t)=\frac{1}{3}e^{2t}+\frac{2}{3}e^{-4t}\)

Solve Equation (c)

The equation \(y^{\prime \prime}+y=0\) is a homogeneous differential equation. From the characteristic equation \(m^2+1=0\), we get \(m=\pm i\). Hence, the general solution is \(y(t)=C_1cos(t)+C_2sin(t)\). Apply the initial conditions \(y(0)=1\) gives \(C_1= 1\). Also, \(y(\pi)=0\), gives \(C_2=0\). Therefore, the solution is \(y(t)=cos(t)\)