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3 years ago

Solve (D*D + 4)y = x + cos²x

Mathematics

1 answer:

Sedaia [141]3 years ago

7 0

Answer:

(D^2 + 9)y = cos 2x….(1). The corresponding homogeneous equation is (D^2 +9)y= 0,…(2), whose auxiliary equation is m^2 + 9 = 0, which has (+/-)3i as roots. The general solution of (2) is y = A.cos(3x) + B.sin(3x). Now to get a general solution of (1) we have just to add to the above, a particular solution of (1). One such solution is [cos(2x)]/[-2^2 +9] = (1/5).cos 2x. Hence a general solution of the given equation is given by y = A.cos(3x) + B.sin(3x) + (1/5)cos(2x), where A and B are arbitrary constants. The above solution incorporates all the solutions of the given equation.

Step-by-step explanation:

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3 years ago

Write an equation for each question, please! (t = years after 1960)

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Answer:

1) y = - 0.54(25)

2) 0.4 = 98 - 0.54t

3) 0.6 = 2 + 0.54t

4) 98 - 0.54t = 2(2 + 0.54t)

(you asked for equations, so I think these are the equations you are looking for!)

Step-by-step explanation:

(given) y = 98 - 0.54t

(plugging in years after 1960) y = - 0.54(25)

(given) y = 98 - 0.54t

(changing percent to decimal) 40% = 0.4

(plugging in percentage) 0.4 = 98 - 0.54t

(given) y = 2 + 0.54t

(changing percent to decimal) 60% = 0.6

(plugging in percentage) 0.6 = 2 + 0.54t

(given)

(male) y = 98 - 0.54t

(female) y = 2 + 0.54t

(setting up the equation so male is 2 times the percent of female)

98 - 0.54t = 2(2 + 0.54t)

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3 years ago

Solve (D*D + 4)y = x + cos²x​

Solve (D*D + 4)y = x + cos²x