y" – 7y" + 25y – 175y = A fundamental set of solutions of the homogeneous equation is given sec 5t, y(0) = 2, y'(0) = -2, y"(0) = 61. by the functions: Y1(t) = eat, where a = %3D Y2(t) Y3(t) : A particular solution is given by: Y (t) = ds y1(t) to · Y2(t) Therefore the solution of the initial-value problem is: y(t) +Y(t) +
y" – 7y" + 25y – 175y = A fundamental set of solutions of the homogeneous equation is given sec 5t, y(0) = 2, y'(0) = -2, y"(0) = 61. by the functions: Y1(t) = eat, where a = %3D Y2(t) Y3(t) : A particular solution is given by: Y (t) = ds y1(t) to · Y2(t) Therefore the solution of the initial-value problem is: y(t) +Y(t) +
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
![```markdown
Given the differential equation:
\[
y''' - 7y'' + 25y' - 175y = \sec 5t, \quad y(0) = 2, \quad y'(0) = -2, \quad y''(0) = 61.
\]
A fundamental set of solutions of the homogeneous equation is given by the functions:
\[
y_1(t) = e^{at}, \quad \text{where } a = \boxed{\phantom{a}}
\]
\[
y_2(t) = \boxed{\phantom{y_2(t)}}
\]
\[
y_3(t) = \boxed{\phantom{y_3(t)}}
\]
A particular solution is given by:
\[
Y(t) = \int_{t_0}^{t} \boxed{\phantom{\int}} \, ds \cdot y_1(t)
\]
\[
\quad \quad + \left(\boxed{\phantom{y_2}}\right) \cdot y_2(t)
\]
\[
\quad \quad + \left(\boxed{\phantom{y_3}}\right) \cdot y_3(t)
\]
Therefore, the solution of the initial-value problem is:
\[
y(t) = \boxed{\phantom{y(t)}} + Y(t)
\]
```
Note: This text involves a differential equation and requires finding specific functions and constants, denoted by boxes, to solve the equation. The integral represents a particular solution to the non-homogeneous part of the differential equation, and the complete solution combines this with the homogeneous solution.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F0bff7fb0-7afb-4d12-af99-7cba91505041%2Ffe9175e0-d5ad-4a49-9252-81321c1fcb15%2Fdl241e_processed.jpeg&w=3840&q=75)
Transcribed Image Text:```markdown
Given the differential equation:
\[
y''' - 7y'' + 25y' - 175y = \sec 5t, \quad y(0) = 2, \quad y'(0) = -2, \quad y''(0) = 61.
\]
A fundamental set of solutions of the homogeneous equation is given by the functions:
\[
y_1(t) = e^{at}, \quad \text{where } a = \boxed{\phantom{a}}
\]
\[
y_2(t) = \boxed{\phantom{y_2(t)}}
\]
\[
y_3(t) = \boxed{\phantom{y_3(t)}}
\]
A particular solution is given by:
\[
Y(t) = \int_{t_0}^{t} \boxed{\phantom{\int}} \, ds \cdot y_1(t)
\]
\[
\quad \quad + \left(\boxed{\phantom{y_2}}\right) \cdot y_2(t)
\]
\[
\quad \quad + \left(\boxed{\phantom{y_3}}\right) \cdot y_3(t)
\]
Therefore, the solution of the initial-value problem is:
\[
y(t) = \boxed{\phantom{y(t)}} + Y(t)
\]
```
Note: This text involves a differential equation and requires finding specific functions and constants, denoted by boxes, to solve the equation. The integral represents a particular solution to the non-homogeneous part of the differential equation, and the complete solution combines this with the homogeneous solution.
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