Question

Let $x_0$ be an arbitrary real number. Given (Example 5.1.1) that $e^x$ and $e^{-x}$ are solutions of $y^{\prime \prime }-y=$ $0,$ find solutions $y_1$ and $y_2$ of $y^{\prime \prime }-y=0$ such that $$y_1\left(x_0\right)=1,y_1^{\prime }\left(x_0\right)=0\text{ and }{y}_2\left(x_0\right)=0,y_2^{\prime }\left(x_0\right)=1$$ Then use Exercise 37 (c) to write the solution of the initial value problem $$y^{\prime \prime }-y=0,y\left(x_0\right)=k_0,y^{\prime }\left(x_0\right)=k_1$$ as a linear combination of $y_1$ and $y_2$.

Short answer

Question: Find the required solutions, $y_1(x)$ and $y_2(x)$, and write the general solution for the given initial value problem: $y^{″}-y=0$, with $y(x_0)=k_0$ and $y^{\prime }(x_0)=k_1$. Answer: The required solutions are $y_1(x)=\frac{1}{2}{e}^x+\frac{1}{2}{e}^{-x}$ and $y_2(x)=\frac{1}{2}{e}^x-\frac{1}{2}{e}^{-x}$. The general solution for the initial value problem is $y(x)=k_0(\frac{1}{2}{e}^x+\frac{1}{2}{e}^{-x})+k_1(\frac{1}{2}{e}^x-\frac{1}{2}{e}^{-x})$.

Step-by-step solution

Find a linear combination of the given solutions

Since we know that $e^x$ and $e^{-x}$ are solutions to the differential equation, we can assume that our required solutions $y_1(x)$ and $y_2(x)$ can be written as a linear combination of the given solutions. Let $y_1(x)=A{e}^x+B{e}^{-x}$ and $y_2(x)=C{e}^x+D{e}^{-x}$, where A, B, C, and D are constants to be determined.

Differentiate the linear combinations

In order to find the constants A, B, C, and D, we need to find the derivatives of the linear combinations with respect to x. So, let's find $y_1^{\prime }(x)$ and $y_2^{\prime }(x)$: $$y_1^{\prime }(x)=A{e}^x-B{e}^{-x}$$ $$y_2^{\prime }(x)=C{e}^x-D{e}^{-x}$$

Apply the initial conditions to find the constants

Now we can apply the initial conditions given in the problem to determine the constants A, B, C, and D. For $y_1(x)$ and its derivative, we have: $$y_1(x_0)=A{e}^{x_0}+B{e}^{-x_0}=1$$ $$y_1^{\prime }(x_0)=A{e}^{x_0}-B{e}^{-x_0}=0$$ And for $y_2(x)$ and its derivative, we have: $$y_2(x_0)=C{e}^{x_0}+D{e}^{-x_0}=0$$ $$y_2^{\prime }(x_0)=C{e}^{x_0}-D{e}^{-x_0}=1$$ We can use these equations to solve for A, B, C, and D: A = $\frac{1}{2}$, B = $\frac{1}{2}$, C = $\frac{1}{2}$, and D = $-\frac{1}{2}$

Write the required solutions as a linear combination of the given solutions

Now that we have the constants, we can write the required solutions $y_1(x)$ and $y_2(x)$ as linear combinations of the given solutions $e^x$ and $e^{-x}$: $$y_1(x)=\frac{1}{2}{e}^x+\frac{1}{2}{e}^{-x}$$ $$y_2(x)=\frac{1}{2}{e}^x-\frac{1}{2}{e}^{-x}$$

Write the general solution of the initial value problem

Finally, we can use Exercise 37(c) to write the solution of the initial value problem as a linear combination of the solutions $y_1(x)$ and $y_2(x)$. The general solution is given by: $$y(x)=k_0{y}_1(x)+k_1{y}_2(x)=k_0(\frac{1}{2}{e}^x+\frac{1}{2}{e}^{-x})+k_1(\frac{1}{2}{e}^x-\frac{1}{2}{e}^{-x})$$