Y''+4y=t^(2)+3e^t find the general solution of the nonhomogenous differential equation
1 answer:
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<h3>Answer: Largest value is a = 9</h3>
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Work Shown:
b = 5
(2b)^2 = (2*5)^2 = 100
So we want the expression a^2+3b to be less than (2b)^2 = 100
We need to solve a^2 + 3b < 100 which turns into
a^2 + 3b < 100
a^2 + 3(5) < 100
a^2 + 15 < 100
after substituting in b = 5.
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Let's isolate 'a'
a^2 + 15 < 100
a^2 < 100-15
a^2 < 85
a < sqrt(85)
a < 9.2195
'a' is an integer, so we round down to the nearest whole number to get
So the greatest integer possible for 'a' is a = 9.
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Check:
plug in a = 9 and b = 5
a^2 + 3b < 100
9^2 + 3(5) < 100
81 + 15 < 100
96 < 100 .... true statement
now try a = 10 and b = 5
a^2 + 3b < 100
10^2 + 3(5) < 100
100 + 15 < 100 ... you can probably already see the issue
115 < 100 ... this is false, so a = 10 doesn't work
Answer:
2x -y ≥ 4
Step-by-step explanation:
The intercepts of the boundary line are given, so it is convenient to start with the equation of that line in intercept form:
... x/(x-intercept) + y/(y-intercept) = 1
... x/2 + y/(-4) = 1
Multiplying by 4 gives the equation of the line.
... 2x -y = 4
This line divides the plane into two half-planes. The half-plane that is shaded is the one for larger values of x and/or smaller values of y than the ones on the line. So, for some given y, if we increase x we will get a number from our equation above that is greater than 4. Hence, the inequality we want is ...
... 2x -y ≥ 4
We use the ≥ symbol because the line is solid, so part of the solution space.
