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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}

Expert verified
The particular solution to the given initial value problem is: $$y(x) = -ee^{-x} + xe^{-x}$$.
See the step by step solution

## 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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