Question

In Exercises $2-7$ solve the boundary value problem. $$y^{\prime \prime }-y=x,y(0)+y^{\prime }(0)=3,y(1)-y^{\prime }(1)=2$$

Short answer

Question: Find the solution to the boundary value problem for the given second-order linear differential equation with boundary conditions: $$y^{\prime \prime }-y=x,y(0)+y^{\prime }(0)=3,y(1)-y^{\prime }(1)=2$$ Answer: The particular solution that satisfies both the differential equation and the boundary conditions is given by: $$y(x)=2e^x+\frac{2}{1-e^{-2}}{e}^{-x}-x$$

Step-by-step solution

Solve the Homogeneous Differential Equation

First, we need to solve the homogeneous differential equation, which is obtained by setting the right-hand side of the given differential equation equal to zero: $$y^{\prime \prime }-y=0$$ The characteristic equation for this ODE is: $$r^2-1=0$$ Factoring the expression, we get $(r-1)(r+1)=0$. Thus, the roots are $r_1=1$ and $r_2=-1$. Therefore, the general solution for the homogeneous differential equation is given by: $$y_h(x)=C_1{e}^x+C_2{e}^{-x}$$ where $C_1$ and $C_2$ are constants that will need to be determined using the boundary conditions.

Find a Particular Solution to the Inhomogeneous Differential Equation

To find a particular solution to the inhomogeneous differential equation, we will use the method of undetermined coefficients. Since the right-hand side of the given differential equation is a polynomial of degree 1 (a linear function), we will assume a particular solution of the form: $$y_p(x)=Ax+B$$ where $A$ and $B$ are constants to be determined. Now, we differentiate $y_p(x)$ twice to get its second derivative: $$y_p^{\prime }(x)=A$$ $$y_p^{\prime \prime }(x)=0$$ Substitute $y_p(x)$, $y_p^{\prime }(x)$, and $y_p^{\prime \prime }(x)$ into the given inhomogeneous differential equation: $$0-(Ax+B)=x$$ Which simplifies to: $$-Ax-B=x$$ Comparing coefficients, we have: $$A=-1$$ and $$B=0$$ Thus, the particular solution to the inhomogeneous differential equation is: $$y_p(x)=-x$$

Combine the Homogeneous and Particular Solutions

Now, we need to combine the homogeneous and particular solutions to obtain the general solution to the given inhomogeneous differential equation: $$y(x)=y_h(x)+y_p(x)=C_1{e}^x+C_2{e}^{-x}-x$$

Apply the Boundary Conditions to Determine the Constants

Now, we will apply the given boundary conditions to determine the values of $C_1$ and $C_2$. First boundary condition is: $$y(0)+y^{\prime }(0)=3$$ Calculate $y^{\prime }(x)$: $$y^{\prime }(x)=C_1{e}^x-C_2{e}^{-x}-1$$ Evaluate the first boundary condition: $$(C_1+C_2)+(C_1-C_2-1)=3$$ Simplifying the equation gives: $$2C_1-1=3$$ From this, we can find the value of $C_1$: $$C_1=2$$ The second boundary condition is: $$y(1)-y^{\prime }(1)=2$$ Plug in the values of $x=1$, and $C_1=2$ into the general solution and its derivative, then apply the second boundary condition: $$(2e+C_2{e}^{-1}-1)-(2e-C_2{e}^{-1}-1)=2$$ Simplifying the equation gives: $$C_2(1-e^{-2})=2$$ We can find the value of $C_2$: $$C_2=\frac{2}{1-e^{-2}}$$

Write the Final Solution

Now, we have the values of $C_1$ and $C_2$. Plug them into the general solution to get the particular solution that satisfies both the differential equation and the boundary conditions: $$y(x)=2e^x+\frac{2}{1-e^{-2}}{e}^{-x}-x$$ This is the final solution to the given boundary value problem.

Key concepts

Understanding Homogeneous Differential Equations

A homogeneous differential equation is a type of differential equation where all the terms are a function of the dependent variable and its derivatives only. In the given exercise, the homogeneous part is identified by setting the right-hand side of the original differential equation to zero.

The equation then simplifies to $y^{″}-y=0$, which only involves the function $y$ and its second derivative $y^{″}$. This equation is considered homogeneous because there are no terms involving only $x$ or constants.

To solve this, we derive the characteristic equation, $r^2-1=0$, from the coefficients of $y$ and its derivatives. This equation helps in finding the roots which are critical in constructing the general solution to the homogeneous differential equation. Here, the roots are $r_1=1$ and $r_2=-1$, leading us to the general solution $y_h(x)=C_1{e}^x+C_2{e}^{-x}$, where $C_1$ and $C_2$ are constants determined by the boundary conditions.

Finding a Particular Solution

When dealing with inhomogeneous differential equations like $y^{″}-y=x$, we seek a particular solution that works for the non-homogeneous part, in this case $x$. The method of undetermined coefficients is a technique where we guess a form for the particular solution based on the type of the non-homogeneous term.

For our equation, since the non-homogeneous term is a polynomial of degree 1 (a linear term $x$), we assume a particular solution of the form $y_p(x)=Ax+B$, where $A$ and $B$ are constants. By substituting $y_p$ and its derivatives back into the original equation and simplifying, we determine the values of the constants.

In this case, $A=-1$ and $B=0$, leading to the particular solution $y_p(x)=-x$. This solution addresses the specific non-homogeneous part of our differential equation and is combined with the homogeneous solution to form the more complete general solution.

Characteristic Equation and Its Role

The characteristic equation is a foundational step in solving homogeneous linear differential equations with constant coefficients. It translates the differential equation into an algebraic form, which allows us to find the roots that represent the solutions to the differential equation.

This algebraic equation is obtained by assuming a solution of the form $y=e^{rx}$ and substituting it into the homogeneous differential equation. For the exercise in question, this leads to the characteristic equation $r^2-1=0$, which then factors into $(r-1)(r+1)=0$ revealing the roots $r_1=1$ and $r_2=-1$.

The roots of the characteristic equation tell us the form of the general solution to the homogeneous differential equation, which is a sum of solutions derived from each root, herein $y_h(x)=C_1{e}^x+C_2{e}^{-x}$. This process shows how the characteristic equation bridges the gap between differential and algebraic equations, making it easier to solve the former by using techniques from the latter.