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Ilia_Sergeevich [38]
4 years ago
8

Need help please. Thanks so much.

Mathematics
1 answer:
Bingel [31]4 years ago
8 0

Answer:

(-1,\frac{17}{6})

(3,\frac{3}{2})

(-5,\frac{25}{6})

Step-by-step explanation:

The given equation is 2x+6y=15

We substitute the ordered pairs to see which ones satisfies the equation.

For (-1,-8), we put x=-1 and y=-8 to get:

2(-1)+6(-8)=^?15

-2+-48=^?15\\-50=^?15

This is not true.

For (-1,\frac{17}{6})) we put x=-1 and y=\frac{17}{6} to get

2(-1)+6*\frac{17}{6})=15\\ -2+17=15\\15=15

This is true

For (3,3/2) we have 2(3)+6(3/2)=15

This implies 6+9=15

15=15

This is also True

For (-5,25/6), we have 2(-5)+6(25/6)=-10+25=15

This is also true.

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The general form of a solution of the differential equation is already provided for us:

y(x) = c_1 \textrm{e}^x + c_2\textrm{e}^{-x},

where c_1, c_2 \in \mathbb{R}. We now want to find a solution y such that y(-1)=3 and y'(-1)=-3. Therefore, all we need to do is find the constants c_1 and c_2 that satisfy the initial conditions. For the first condition, we have:y(-1)=3 \iff c_1 \textrm{e}^{-1} + c_2 \textrm{e}^{-(-1)} = 3 \iff c_1\textrm{e}^{-1} + c_2\textrm{e} = 3.

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If we now substitute c_1 = 0 into either of the equations in the system, we get:

c_2 \textrm{e} = 3 \iff c_2 = \dfrac{3}{\textrm{e}} = 3\textrm{e}^{-1.}

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\boxed{y(x) = 3\textrm{e}^{-1} \times \textrm{e}^{-x} = 3\textrm{e}^{-x-1}}.

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y(-1) = 3\textrm{e}^{-(-1) -1} = 3\textrm{e}^{1-1} = 3\textrm{e}^0 = 3

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which do correspond to the desired initial conditions.

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