Solve the initial value problem: \( y^{\prime}(x)=\sqrt{4 y(x)+17}, \quad y(-2)=2 \) \( y(x)= \) Note: observe that the \( y(x) \) under the square root is not y times x . It is function notation, i.e. " y of x ".
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To solve the initial value problem \( y^{\prime}(x) = \sqrt{4 y(x) + 17} \) with the initial condition \( y(-2) = 2 \), we start by separating the variables. 1. Rewrite the equation: \[ \frac{dy}{dx} = \sqrt{4y + 17} \] 2. Separate the variables \( y \) and \( x \): \[ \frac{dy}{\sqrt{4y + 17}} = dx \] 3. Now, let's integrate both sides. Starting with the left side: To integrate the left side, we can use the substitution method. Let \( u = 4y + 17 \), then \( du = 4 dy \) or \( dy = \frac{du}{4} \). Thus, \[ \int \frac{1}{\sqrt{u}} \cdot \frac{1}{4} du = \frac{1}{4} \cdot 2\sqrt{u} + C = \frac{1}{2} \sqrt{4y + 17} + C \] And the right side is simply: \[ \int dx = x + C_1 \] 4. Putting this together, we have: \[ \frac{1}{2} \sqrt{4y + 17} = x + C \] 5. Multiplying both sides by 2: \[ \sqrt{4y + 17} = 2x + 2C \] 6. Squaring both sides gives: \[ 4y + 17 = (2x + 2C)^2 \] 7. Simplifying leads to: \[ 4y = 4x^2 + 8Cx + 4C^2 - 17 \] \[ y = x^2 + 2Cx + C^2 - \frac{17}{4} \] 8. Now, we will use the initial condition \( y(-2) = 2 \) to find \( C \): \[ 2 = (-2)^2 + 2C(-2) + C^2 - \frac{17}{4} \] \[ 2 = 4 - 4C + C^2 - \frac{17}{4} \] This simplifies to: \[ 2 = \frac{16}{4} - 4C + C^2 - \frac{17}{4} \] \[ 2 = -4C + C^2 - \frac{1}{4} \] Multiply through by 4 to eliminate the fraction: \[ 8 = -16C + 4C^2 - 1 \] \[ 4C^2 - 16C - 9 = 0 \] Using the quadratic formula, \( C = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \): \[ C = \frac{16 \pm \sqrt{256 + 144}}{8} = \frac{16 \pm 20}{8} \] So, \( C = 4.5 \) or \( C = -0.5 \). 9. We would take the appropriate \( C \) from the context of the problem (usually the positive constant would be preferred unless a particular behavior is needed). 10. Assuming \( C = -0.5 \), we have: \[ y = x^2 - (2)(0.5)x + 0.25 - \frac{17}{4} \] Thus upon simplifying yields: \[ y = x^2 - x - \frac{15}{4} \] Therefore, the solution to the initial value problem is: \[ y(x) = x^2 - x - \frac{15}{4} \]