Given that y₁ (t) = cost is a solution to y'' -y' + y = sint and y₂ (t)=3 is a solution to y'' - y' + y = e²t, use the superposition principle to find solutions to the differential equations in parts (a) through (c) below. (a) y'-y'+y=12 sin t A solution is y(t) = (b) y'-y'+y=3 sint-6 e 2t A solution is y(t) =

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
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Chapter2: Second-order Linear Odes
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part a and b

Given that \( y_1(t) = \cos t \) is a solution to \( y'' - y' + y = \sin t \) and \( y_2(t) = \frac{e^{2t}}{3} \) is a solution to \( y'' - y' + y = e^{2t} \), use the superposition principle to find solutions to the differential equations in parts (a) through (c) below.

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**(a)** \( y'' - y' + y = 12 \sin t \)  
A solution is \( y(t) = \boxed{\phantom{y(t)}} \).

**(b)** \( y'' - y' + y = 3 \sin t - 6 e^{2t} \)  
A solution is \( y(t) = \boxed{\phantom{y(t)}} \).

**(c)** \( y'' - y' + y = 5 \sin t + 15 e^{2t} \)  
A solution is \( y(t) = \boxed{\phantom{y(t)}} \).
Transcribed Image Text:Given that \( y_1(t) = \cos t \) is a solution to \( y'' - y' + y = \sin t \) and \( y_2(t) = \frac{e^{2t}}{3} \) is a solution to \( y'' - y' + y = e^{2t} \), use the superposition principle to find solutions to the differential equations in parts (a) through (c) below. --- **(a)** \( y'' - y' + y = 12 \sin t \) A solution is \( y(t) = \boxed{\phantom{y(t)}} \). **(b)** \( y'' - y' + y = 3 \sin t - 6 e^{2t} \) A solution is \( y(t) = \boxed{\phantom{y(t)}} \). **(c)** \( y'' - y' + y = 5 \sin t + 15 e^{2t} \) A solution is \( y(t) = \boxed{\phantom{y(t)}} \).
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