This is algebra 1 id appreciate it if you could help me thx!

Mathematics

1 answer:

Mnenie [13.5K]3 years ago

7 0

Answer:

Option 4) y - 1 = -2(x - 0)

Step-by-step explanation:

Given the slope, m = -2, and the y-intercept, (0, 1):

Substitute these values into the following point-slope form:

y - y1 = m(x - x1)

y - 1 = -2(x - 0) ← This is the point-slope form given the slope, m = -2, and the y-intercept, (0, 1).

Therefore, the correct answer is Option 4: y - 1 = -2(x - 0)

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Y''-2y'+y=2(e^x)-3(e^-x)

labwork [276]

First find the characteristic solution. The characteristic equation is

$r^2-2r+1=(r-1)^2=0$

which as one root at $r=1$ of multiplicity 2. This means the characteristic solution for this ODE is

$y_c=C_1e^x+C_2xe^x$

For the nonhomogeneous part, you can try a particular solution of the form

$y_p=(a_2x^2+a_1x+a_0)e^x+be^{-x}$

which has derivatives

${y_p}'=(a_2x^2+(2a_2+a_1)x+a_1+a_0)e^x-be^{-x}$
${y_p}''=(a_2x^2+(4a_2+a_1)x+2a_2+2a_1+a_0)e^x+be^{-x}$

Substituting into the ODE, the left hand side reduces significantly to

$2a_2e^x+4be^{-x}=2e^x-3e^{-x}$

and it follows that

$\begin{cases}2a_2=2\\4b=-3\end{cases}\implies a_2=1,b=-\dfrac34$

Therefore the particular solution is

$y_p=x^2e^x-\dfrac34e^{-x}$

and so the general solution is the sum of the characteristic and particular solutions,

$y=y_c+y_p$
$y=C_1e^x+C_2xe^x+x^2e^x-\dfrac34e^{-x}$