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Find the general solution of the given differential equation. y"-14y +49y=0

Find the general solution of the given differential equation. y"-14y +49y=0

Calculus For The Life Sciences
Calculus For The Life Sciences
2nd Edition
ISBN:9780321964038
Author:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Publisher:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Chapter11: Differential Equations
Section11.1: Solutions Of Elementary And Separable Differential Equations
Problem 17E: Find the general solution for each differential equation. Verify that each solution satisfies the...
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**Topic: Solutions of Differential Equations** --- **Problem Statement:** Find the general solution of the given differential equation. \[ y'' - 14y' + 49y = 0 \] --- **Explanation:** In this section, we will solve the given second-order linear homogeneous differential equation with constant coefficients. The equation provided is: \[ y'' - 14y' + 49y = 0 \] **Steps to Solve:** 1. **Form the Characteristic Equation:** Convert the given differential equation to its corresponding characteristic (auxiliary) equation by replacing \( y'' \) with \( r^2 \), \( y' \) with \( r \), and \( y \) with 1. \[ r^2 - 14r + 49 = 0 \] 2. **Solve the Characteristic Equation:** Solve the quadratic equation for \( r \). The characteristic equation can be simplified as: \[ (r - 7)^2 = 0 \] This indicates a repeated real root \( r = 7 \). 3. **Form the General Solution:** For a repeated real root \( r = 7 \), the general solution of the differential equation is given by: \[ y(x) = (C_1 + C_2 x) e^{7x} \] where \( C_1 \) and \( C_2 \) are arbitrary constants determined by initial conditions or boundary values. Hence, the general solution of the differential equation is: \[ y(x) = (C_1 + C_2 x) e^{7x} \] --- *Note:* For further practice, try solving differential equations with distinct roots or complex roots and observe how the solution form changes accordingly.
Transcribed Image Text:**Topic: Solutions of Differential Equations** --- **Problem Statement:** Find the general solution of the given differential equation. \[ y'' - 14y' + 49y = 0 \] --- **Explanation:** In this section, we will solve the given second-order linear homogeneous differential equation with constant coefficients. The equation provided is: \[ y'' - 14y' + 49y = 0 \] **Steps to Solve:** 1. **Form the Characteristic Equation:** Convert the given differential equation to its corresponding characteristic (auxiliary) equation by replacing \( y'' \) with \( r^2 \), \( y' \) with \( r \), and \( y \) with 1. \[ r^2 - 14r + 49 = 0 \] 2. **Solve the Characteristic Equation:** Solve the quadratic equation for \( r \). The characteristic equation can be simplified as: \[ (r - 7)^2 = 0 \] This indicates a repeated real root \( r = 7 \). 3. **Form the General Solution:** For a repeated real root \( r = 7 \), the general solution of the differential equation is given by: \[ y(x) = (C_1 + C_2 x) e^{7x} \] where \( C_1 \) and \( C_2 \) are arbitrary constants determined by initial conditions or boundary values. Hence, the general solution of the differential equation is: \[ y(x) = (C_1 + C_2 x) e^{7x} \] --- *Note:* For further practice, try solving differential equations with distinct roots or complex roots and observe how the solution form changes accordingly.
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