Consider the following pair of equations.

Mathematics

1 answer:

RUDIKE [14]2 years ago

4 0

Answer:

(- 2, 2 )

Step-by-step explanation:

Given the 2 equations

y = x + 4 → (1)

y = - 2x - 2 → (2)

Substitute y = x + 4 into (2)

x + 4 = - 2x - 2 ( add 2x to both sides )

3x + 4 = - 2 ( subtract 4 from both sides )

3x = - 6 ( divide both sides by 3 )

x = - 2

Substitute x = - 2 into either of the 2 equations abd evaluate for y

Substituting into (1)

y = - 2 + 4 = 2

solution is (- 2, 2 )

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A group of boys picked 216 boxes of oranges. each box contained 120 oranges.how many oranges did they pick?

miv72 [106K]

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Step-by-step explanation:

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3 0

2 years ago

Y"-4y'+4y=0 y1=e^2x find the general solution

Karo-lina-s [1.5K]

Given that $y_1=e^{2x}$ is a known solution to the ODE, we can reduce the order of the ODE to find a second solution of the form

$y_2=y_1v=e^{2x}v$
$\implies {y_2}'=2e^{2x}v+e^{2x}v'$
$\implies {y_2}''=4e^{2x}v+4e^{2x}v'+e^{2x}v''$

Substituting into the ODE, we get

$(e^{2x}v''+4e^{2x}v'+4e^{2x}v)-4(e^{2x}v'+2e^{2x}v)+4e^{2x}v=e^{2x}$
$e^{2x}v''=e^{2x}$
$v''=1$
$\implies v'=C_1$
$\implies v=C_1x+C_2$
$\implies y_2=C_1xe^{2x}+C_2e^{2x}$

We already know about $e^{2x}$ as a solution to the ODE, which means $y_2=xe^{2x}$ .

That covers the characteristic solution. To find the particular solution to the nonhomogeneous ODE, suppose there is a solution of the form

$y_p=ax^2e^{2x}$
${y_p}'=2ax(x+1)e^{2x}$
${y_p}''=2a(2x^2+4x+1)e^{2x}$

Substituting into the ODE yields

${y_p}''-4{y_p}'+4{y_p}=2ae^{2x}=e^{2x}\implies a=\dfrac12$

so that the general solution is

$y=C_1y_1+C_2y_2+y_p$
$y=C_1e^{2x}+C_2xe^{2x}+\dfrac12x^2e^{2x}$