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rosijanka [135]
3 years ago
7

Consider the homogeneous second-order linear differential equation y′′+4y′−12y=0. Which of the following pairs gives two solutio

ns to this equation? A. y1=e2x,y2=e−6x B. y1=e3x,y2=e1x C. y1=e2x,y2=e−2x D. y1=e−12x,y2=xe−12x E. y1=cos(−12x),y2=sin(−12x) F. y1=e−4x,y2=e−12x Then for these solutions find a particular solution of the form y=c1y1+c2y2 that satisfies the initial conditions y(0)=−5,y′(0)=0. y = y1 + y2.
Mathematics
1 answer:
SOVA2 [1]3 years ago
3 0
Cc’ing chi c gg hi barb cub
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Answer:

Simplifying

5x + 12 = -13

Reorder the terms:

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12 + 5x = -13

Solving for variable 'x'.

Move all terms containing x to the left, all other terms to the right.

Add '-12' to each side of the equation.

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Combine like terms: 12 + -12 = 0

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Step-by-step explanation:

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