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2 years ago

What is the value of y in the solution to the following system of equations?

Mathematics

1 answer:

lukranit [14]2 years ago

4 0

Answer:

Option D is the correct answer.

Step-by-step explanation:

The equations are:

5x - 3y = -11 Eqn 1

2x - 6y = -14 Eqn 2

We have to find the value of y so we will eliminate x.

Multiplying Eqn 1 by 2

2(5x-3y=-11)

10x-6y=-22 Eqn 3

Multiplying Eqn 2 by 5

5(2x-6y=-14)

10x-30y=-70 Eqn 4

Subtracting Eqn 4 from Eqn 3

(10x-6y)-(10x-30y)= -22-(-70)

10x-6y-10x+30y = -22 + 70

24y = 48

$\frac{24y}{24}=\frac{48}{24}\\y=2$

Therefore,

Option D is the correct answer.

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Find a particular solution to the nonhomogeneous differential equation y??+4y?+5y=?10x+e^(?x).

Firdavs [7]

Answer:

A) Particular solution:

$2x+\frac{1}{2}e^{-x}-\frac{8}{5}$

B) Homogeneous solution:

$y_{h}=e^{-2x}(c_{1}cos(x)+c_{2}sin(x))$

C) The most general solution is

$y=e^{-2x}(c_{1}cos(x)+c_{2}sin(x))+2x+\frac{1}{2}e^{-x}-\frac{8}{5}$

Step-by-step explanation:

Given non homogeneous ODE is

$y''+4y'+5y=10x+e^{-x}---(1)$

To find homogeneous solution:

$D^{2}+4D+5=0\\D^{2}+4D+4-4+5=0\\\\(D+2)^{2}=-1\\D+2=\pm iD=-2 \pm i\\y_{h}=e^{-2x}(c_{1}cos(x)+c_{2}sin(x))---(2)$

To find particular solution:

$y_{p}=Ax+B+Ce^{-x}\\\\y'_{p}=A-Ce^{-x}\\y''_{p}=Ce^{-x}\\$

Substituting $y_{p},y'_{p},y''_{p}$ in (1)

$y''_{p}+4y'_{p}+5y_{p}=10x+e^{-x}\\Ce^{-x}+4(A-Ce^{-x})+5(Ax+B+Ce^{-x})=10x+e^{-x}\\$

Equating the coefficients

$5Ax+2Ce^{-x}+4A+5B=10x+e^{-x}\\5A=10\\A=2\\4A+5B=0\\B=-\frac{4A}{5}B=-\frac{8}{5}2C=1\\C=\frac{1}{2}\\So,\\y_{p}=2x+\frac{1}{2}e^{-x}-\frac{8}{5}---(3)\\$

The general solution is

$y=y_{h}+y_{p}$

from (2) ad (3)

$y=e^{-2x}(c_{1}cos(x)+c_{2}sin(x))+2x+\frac{1}{2}e^{-x}-\frac{8}{5}$

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Answer:

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Step-by-step explanation:

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$538.92 - 10 = 528.92$

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