Answer:
y=2e^(−x)cosx−e^(−x)sinx
Satisfies the equation
Step-by-step explanation:
Answer:
y=2e^(−x)cosx−e^(−x)sinx
y = e^(-x)[2cosx - sinx]
Find y' and y" using product law
y' = -e^(-x)[2cosx - sinx] + e^(-x)[-2sinx - cosx]
y' = -e^(-x)[2cosx - sinx + 2sinx + cosx]
y' = -e^(-x)[3cosx + sinx]
y" = e^(-x)[3cosx + sinx] - e^(-x)[-3sinx + cosx]
y" = e^(-x)[3cosx - cosx + sinx + 3sinx]
y" = e^(-x)[2cosx + 4sinx]
y" + 2y' + 2y
e^(-x)[2cosx + 4sinx] - 2e^(-x)[3cosx + sinx] +2e^(-x)[2cosx - sinx]
e^(-x)[4sinx - 2sinx - 2sinx + 2cosx - 6 cosx + 4cosx]
= e^(-x) × 0
= 0
I hope this is the answer:)
3x-x+2=4
sorry... this answer is VERY wrong
Answer:
0
Step-by-step explanation:
This is a horizontal line. The equation of the line is y = 10
