Someone please help marking brainliest

Mathematics

1 answer:

Luda [366]3 years ago

3 0

Answer: (1, 4)

Explanation: When using the method of elimination, the goal is to eliminate a variable by either adding or subtracting the 2 equations. For this question, you can choose either to eliminate X or Y. I’ll eliminate X as an example:

In order to eliminate a variable, the same variable in both equations must have the same coefficient.

(1) 3x+y=7
(2) 2x+5y=22

Multiply (1) by 2:

(3) 6x+2y=14

Multiply (2) by 3:

(4) 6x+15y=66

Now that X in both equations has the same coefficient of 6, you can subtract the two equations to officially eliminate the variable and solve for Y:

Subtract (4) from (3):

-13y=-52
y=4

Now that you have the value of Y, substitute that into either one of the equations to get X. I’ll use the first equation as an example:

3x+(4)=7
3x=3
x=1

Therefore, the point of intersection is (1, 4).

Hope this helps シ

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A) Particular solution:

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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

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