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Problem 428

Solve the following initial value problem: $$ \begin{aligned} \mathrm{y}^{\prime \prime}+2 \mathrm{y}^{\prime}+\mathrm{y} &=0 \\ \mathrm{y}(1) &=0 \\ \mathrm{y}^{\prime}(1) &=2 . \end{aligned} $$

Short Answer

Expert verified
The particular solution to the given initial value problem is: \(y(x) = -ee^{-x} + xe^{-x}\).
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Step 1: Solve for the characteristic equation

To find the characteristic equation, we replace \(y^{\prime \prime}\) with \(m^2\), \(y^{\prime}\) with \(m\), and \(y\) with 1. This gives us: \[m^2 + 2m + 1 = 0\]

Step 2: Find the general solution based on the roots of the characteristic equation

Now, we need to solve the quadratic equation for \(m\), obtained in step 1. This equation has a repeated root at \(m = -1\), since \((m + 1)^2 = 0\). So, the general solution of the given differential equation takes the form: \[y(x) = C_1e^{-x} + C_2xe^{-x}\]

Step 3: Apply the initial conditions to find the particular solution

We are given the following initial conditions: \[y(1) = 0\] \[y^{\prime}(1) = 2\] First, apply the initial condition \(y(1) = 0\): \[0 = C_1e^{-1} + C_2(1)e^{-1}\] \[0 = C_1e^{-1} + C_2e^{-1}\] Now, we need to find the derivative \(y^{\prime}(x)\) and apply the initial condition \(y^{\prime}(1) = 2\): \[y^{\prime}(x) = -C_1e^{-x} + C_2e^{-x} - C_2xe^{-x}\] Now, apply the initial condition \(y^{\prime}(1) = 2\): \[2 = -C_1e^{-1} + C_2e^{-1} - C_2e^{-1}\] From the initial conditions equations, we can solve for \(C_1\) and \(C_2\): \[C_1e^{-1} = -C_2e^{-1}\] \[C_1 = -C_2\] Substitute this in the second initial condition equation: \[2 = 2C_2e^{-1}\] Now, solve for \(C_2\): \[C_2 = e\] From the equation \(C_1 = -C_2\), we can find \(C_1\): \[C_1 = -e\] Thus, the particular solution is: \[y(x) = -ee^{-x} + exe^{-x}\]

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Most popular questions from this chapter

Chapter 18

Solve the following initial value problem: $$ \begin{aligned} &x^{\prime}=4 x+y \\ &y^{\prime}=3 x+2 y \end{aligned} $$ Initial conditions: \(\quad \mathrm{x}(0)=-1\) $$ \mathrm{y}(0)=7 $$

Chapter 18

Find the general solution of the following system of differential equations: $$ \begin{aligned} &3\\{(d x / d t)+(d y / d t)\\}=\sin t \\ &(d x / d t)+3(d y / d t)=\cos t \end{aligned} $$

Chapter 18

Let $$ \begin{aligned} &(d x / d t)=a x+b y \\ &(d y / d t)=c x+d y . \end{aligned} $$ Discuss the nature of the critical point of \((1)\) when the roots of the associated characteristic equation are: a) complex; b) pure imaginarv

Chapter 18

Solve the following initial value problem: $$ \begin{gathered} \mathrm{y}_{1}^{\prime}=\mathrm{y}_{1}+\mathrm{y}_{2} \\ \mathrm{y}_{2}^{\prime}=4 \mathrm{y}_{1}-2 \mathrm{y}_{2} \\ \text { Initial conditions: } \mathrm{y}_{1}(0)=1, \mathrm{y}_{2}(0)=6 \end{gathered} $$

Chapter 18

The general solution of the system $$ \begin{aligned} &\mathrm{d} \mathrm{x} / \mathrm{dt}=\mathrm{ax}+\mathrm{by} \\ &\mathrm{dy} / \mathrm{dt}=\mathrm{cx}+\mathrm{dy} \end{aligned} $$ is $$ \mathrm{x}=\mathrm{A}_{1} \mathrm{e}^{(\mathrm{r}) 1(\mathrm{t})}+\mathrm{A}_{2} \mathrm{e}^{(\mathrm{r}) 2(\mathrm{t})}, \mathrm{y}=\mathrm{B}_{1} \mathrm{e}^{(\mathrm{r}) 1(\mathrm{t})}+\mathrm{B}_{2} \mathrm{e}^{(\mathrm{r}) 2(\mathrm{t})} $$ Discuss, with the aid of examples, the nature of the critical point when: a) \(\mathrm{r}_{1}\) and \(\mathrm{r}_{2}\) are real, and of the same sign, but \(\mathrm{r}_{1} \neq \mathrm{r}_{2}\); b) \(\mathrm{r}_{1}\) and \(\mathrm{r}_{2}\) are real, and of opposite signs, and \(\mathrm{r}_{1} \neq \mathrm{r}_{2}\).

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