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4 years ago

Please help me with my algebra.

Mathematics

2 answers:

MA_775_DIABLO [31]4 years ago

6 0

Answer:

Step-by-step explanation:

aalyn [17]4 years ago

3 0

Answer:

Step-by-step explanation:

x+2y=3

5x-3y=2

Subtract 2y from both sides of the first equation to isolate x:

x=3-2y

Substitute this into the second equation:

5(3-2y)-3y=2

15-13y=2

y=1

x=1

Therefore, the correct answer is choice A. Hope this helps!

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3 years ago

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$y_1=e^x\cos x$
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The particular solution is easiest to obtain via variation of parameters. We're looking for a solution of the form

$y_p=u_1y_1+u_2y_2$

where

$u_1=-\displaystyle\frac13\int\frac{y_2e^x\sec x}{W(y_1,y_2)}\,\mathrm dx$
$u_2=\displaystyle\frac13\int\frac{y_1e^x\sec x}{W(y_1,y_2)}\,\mathrm dx$

and $W(y_1,y_2)$ is the Wronskian of the fundamental solutions. We have

$W(e^x\cos x,e^x\sin x)=\begin{vmatrix}e^x\cos x&e^x\sin x\\e^x(\cos x-\sin x)&e^x(\cos x+\sin x)\end{vmatrix}=e^{2x}$

and so

$u_1=-\displaystyle\frac13\int\frac{e^{2x}\sin x\sec x}{e^{2x}}\,\mathrm dx=-\int\tan x\,\mathrm dx$
$u_1=\dfrac13\ln|\cos x|$

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Therefore the particular solution is

$y_p=\dfrac13e^x\cos x\ln|\cos x|+\dfrac13xe^x\sin x$

so that the general solution to the ODE is

$y=C_1e^x\cos x+C_2e^x\sin x+\dfrac13e^x\cos x\ln|\cos x|+\dfrac13xe^x\sin x$

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