1. The differential equation +3 + 2x = 0 has the general solution x(1) = c1e+c2e-2 (by using the characteristic equation). If we are told that, when t = 0, x(0) = 1 and its derivative x'(0) = 1, we can determine c, and cz by solving the equations x(0) = cje- + cz el-2)(0) = c¡(1) + c2(1) = 1, %3D x'(0) = c1(-1)e-0 + c>(-2)e~2x0) = -c1 – 2c2 = 1. (a) Find the values of c and c, by solving the above. (b) Find the solution of the differential equation subject to the specified initial condition by using the above method. i. y" – 3y - 4y = 0; y(0) = 0, y'(0) = 1. ii. y" – 9y' = 0; y(0) = 5, y'(0) = 0. %3D %3D %3D
1. The differential equation +3 + 2x = 0 has the general solution x(1) = c1e+c2e-2 (by using the characteristic equation). If we are told that, when t = 0, x(0) = 1 and its derivative x'(0) = 1, we can determine c, and cz by solving the equations x(0) = cje- + cz el-2)(0) = c¡(1) + c2(1) = 1, %3D x'(0) = c1(-1)e-0 + c>(-2)e~2x0) = -c1 – 2c2 = 1. (a) Find the values of c and c, by solving the above. (b) Find the solution of the differential equation subject to the specified initial condition by using the above method. i. y" – 3y - 4y = 0; y(0) = 0, y'(0) = 1. ii. y" – 9y' = 0; y(0) = 5, y'(0) = 0. %3D %3D %3D
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![**Differential Equations Problem**
1. The differential equation \(\frac{d^2x}{dt^2} + 3 \frac{dx}{dt} + 2x = 0\) has the general solution \(x(t) = c_1 e^{-t} + c_2 e^{-2t}\) (by using the characteristic equation). If we are told that, when \(t = 0\), \(x(0) = 1\) and its derivative \(x'(0) = 1\), we can determine \(c_1\) and \(c_2\) by solving the equations
\[
x(0) = c_1 e^0 + c_2 e^{-2(0)} = c_1(1) + c_2(1) = 1,
\]
\[
x'(0) = c_1(-1)e^0 + c_2(-2) e^{-2(0)} = -c_1 - 2c_2 = 1.
\]
(a) **Find the values of \(c_1\) and \(c_2\) by solving the above.**
(b) **Find the solution of the differential equation subject to the specified initial condition by using the above method.**
i. \(y'' - 3y' - 4y = 0;\quad y(0) = 0,\ y'(0) = 1.\)
ii. \(y'' - 9y' = 0;\quad y(0) = 5,\ y'(0) = 0.\)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F11d9adce-31d3-450e-a738-16d52e01b365%2F1c1c7890-4ea2-4a23-9ed4-650b47b546f5%2Fsgsiuzr_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Differential Equations Problem**
1. The differential equation \(\frac{d^2x}{dt^2} + 3 \frac{dx}{dt} + 2x = 0\) has the general solution \(x(t) = c_1 e^{-t} + c_2 e^{-2t}\) (by using the characteristic equation). If we are told that, when \(t = 0\), \(x(0) = 1\) and its derivative \(x'(0) = 1\), we can determine \(c_1\) and \(c_2\) by solving the equations
\[
x(0) = c_1 e^0 + c_2 e^{-2(0)} = c_1(1) + c_2(1) = 1,
\]
\[
x'(0) = c_1(-1)e^0 + c_2(-2) e^{-2(0)} = -c_1 - 2c_2 = 1.
\]
(a) **Find the values of \(c_1\) and \(c_2\) by solving the above.**
(b) **Find the solution of the differential equation subject to the specified initial condition by using the above method.**
i. \(y'' - 3y' - 4y = 0;\quad y(0) = 0,\ y'(0) = 1.\)
ii. \(y'' - 9y' = 0;\quad y(0) = 5,\ y'(0) = 0.\)
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