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Problem 427

# Find the general solution to each of the following differential equations: $$[1] \mathrm{y}^{\prime}-\mathrm{y}=0 .$$ [2] $$\mathrm{y}^{\prime \prime}+4 \mathrm{y}^{\prime}-\mathrm{y}=0$$. [3] $$\mathrm{y}^{\prime \prime}+8 \mathrm{y}^{\prime}+16 \mathrm{y}=0$$

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The general solutions for the given differential equations are: 1. $$y = Ce^x$$ 2. $$y(x) = A e^{(-2 + \sqrt{5})x} + B e^{(-2 - \sqrt{5})x}$$ 3. $$y(x) = (A + Bx)e^{-4x}$$
See the step by step solution

## Step 1: Solution for equation [1]

To solve the first-order linear ODE, we can use the method of separation of variables. We have: $y' - y = 0$ Rearranging the equation, we get: $y' = y$ Now, we can separate the variables: $\frac{dy}{y} = dx$ Next, we will integrate both sides with respect to their respective variables: $\int \frac{dy}{y} = \int dx$ $\ln |y| = x + C_1$ Now, we will use exponentiation to solve for y: $y = e^{x + C_1}$ $y = e^{C_1}e^x$ Since $$e^{C_1}$$ is an arbitrary constant, we can write the general solution for equation [1] as: $y = Ce^x$

## Step 2: Solution for equation [2]

To solve the second-order linear homogeneous ODE, we will use the method of solving the characteristic equation. We start with the given equation: $y'' + 4y' - y = 0$ The characteristic equation of the given ODE is: $r^2 + 4r - 1 = 0$ Solving the quadratic equation, we get: $r = \frac{-4\pm \sqrt{4^2 - 4(-1)}}{2}$ $r = -2 \pm \sqrt{5}$ As the roots are real and distinct, the general solution for equation [2] is: $y(x) = A e^{(-2 + \sqrt{5})x} + B e^{(-2 - \sqrt{5})x}$

## Step 3: Solution for equation [3]

To solve the second-order linear homogeneous ODE, we will also use the method of solving the characteristic equation. We start with the given equation: $y'' + 8y' + 16y = 0$ The characteristic equation for this ODE is: $r^2 + 8r + 16 = 0$ Solving the quadratic equation, we get: $r = \frac{-8 \pm \sqrt{8^2 - 4(16)}}{2}$ $r = -4$ As the roots are real and repeated, the general solution for equation [3] is: $y(x) = (A + Bx)e^{-4x}$ In conclusion, the general solutions for the given differential equations are: 1. $$y = Ce^x$$ 2. $$y(x) = A e^{(-2 + \sqrt{5})x} + B e^{(-2 - \sqrt{5})x}$$ 3. $$y(x) = (A + Bx)e^{-4x}$$

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