7. Consider the following IVP with discontinuous right hand side. (1 x < 1 0 x ≥1 (c) y = (d) y = y" + 2y + y = The complementary solution to this equation is y₁ = C₁е¯ + c₂xe. Solve this IVP. (a) y = e-+ xe¬ª (b) y=1+xe= (e) y = 1+re x < 1 -² + xе-² x>1 1+re-r x < 1 re-*+re-2+1 x>1 3 e-² + xе-² x < 1 1+re=¹ x>1 y(0) = 1, y'(0) = 1.

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
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7. Consider the following IVP with discontinuous right hand side.
(c) y =
(d) y =
The complementary solution to this equation is yc = c₁e-² + c₂re. Solve this IVP.
(a) y = e. +xe-
(b) y=1+re-
(e) y =
y" + 2y + y =
[1+xe-2
e tre-
x < 1
x>1
1+re-r
xe¯² + xe¬²+1
x < 1
x ≥ 1
x < 1
0 221'
e-tre-e x < 1
1+xe-r
x>1
y(0) = 1, y'(0) = 1.
Transcribed Image Text:7. Consider the following IVP with discontinuous right hand side. (c) y = (d) y = The complementary solution to this equation is yc = c₁e-² + c₂re. Solve this IVP. (a) y = e. +xe- (b) y=1+re- (e) y = y" + 2y + y = [1+xe-2 e tre- x < 1 x>1 1+re-r xe¯² + xe¬²+1 x < 1 x ≥ 1 x < 1 0 221' e-tre-e x < 1 1+xe-r x>1 y(0) = 1, y'(0) = 1.
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