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Dvinal [7]
3 years ago
6

tell whether each equation has one solution infinitely many solutions or no solution justify your answer 3-7t=-5t+3-2t​

Mathematics
1 answer:
NeTakaya3 years ago
4 0

Answer: All real numbers so infinite solutions

Step-by-step explanation:

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I'm stuck on this:
Naily [24]
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3 years ago
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Solve the given initial-value problem. y'' 4y' 5y = 35e−4x, y(0) = −5, y'(0) = 1
meriva

The solution for the initial value problem is y_{g} = e^{-2x} (-12cos(x) + 5sin(x)) + 7e^{-4x}

Given,

y" + 4y' + 5y = 35e^{-4x}

y(0) = -5

y'(0) = 1

Solve this homogenous equation to get y_{h}

According to differential operator theorem,

y_{h} = e^{ax}( A cos (bx) + B sin (bx)), where A and B are constants.

Therefore,

y" + 4y' + 5y = 0

(D^{2} + 4D + 5)y = 0

D = -2± i    

y_{h} = e^{-2x} ( A cos (x) + B sin (x))

Now, solve for y_{p}

A function of the kind ce^{-4x} is the function on the right, we are trying a solution of the form y_{p} =ce^{-4x}, here c is a constant.

y_{p} " + 4y_{p} ' + 5y_{p}  = 35e^{-4x} \\=16ce^{-4x} -16ce^{-4x} +5ce^{-4x} = 35e^{-4x} \\= 5ce^{-4x} =35e^{-4x} \\c=\frac{35}{5} =7\\y_{p} =7e^{-4x}

Then the general solution will be like:

y_{g} =y_{h} +y_{p} \\

    = e^{-2x} (Acos(x)+Bsin(x))+7e^{-4x}

y_{g}(0)=-5=A+7=-12\\y_{g} '(0)=e^{-2x} (-Asin(x)+Bcos(x))-2e^{-2x} (Acos(x)+Bsin(x))-28e^{-4x} \\y'_{g} (0)=1=B-2A-28\\

       B = 1 - 24 + 28 = 5

Then the solution for the given initial value problem is

y_{g} =e^{-2x} (-12cos(x)+5sin(x))+7e^{-4x}

Learn more about initial value problem here: brainly.com/question/8599681

#SPJ4

8 0
2 years ago
Use the formula V = s³, where V is the volume and s is the edge length of the cube, to solve this problem.
Varvara68 [4.7K]
Given that volume is:
V=s³
where:
s=side length
The volume of a cube whose side length, s=3/4 yards will be given by:
V=(3/4)³
Simplifying this we get:
V=27/64 yd³

Answer: 27/64 yd³
8 0
3 years ago
Read 2 more answers
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