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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] \(\mathrm{y}^{\prime \prime}+8 \mathrm{y}^{\prime}+16 \mathrm{y}=0\)

Short Answer

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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. \(y(x) = (A + Bx)e^{-4x}\)
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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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