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4 years ago

13

1st out of 10 questions

1 answer:

Sonja [21]4 years ago

3 0

(4,18)

the slope is 9/2

so (2,9) plus the slope is (4, 18)

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If you're measuring the area of a room in your house, which unit of measurement would be most appropriate?

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3 years ago

Evaluate <br> 2/3b-5a <br> when a=-2 and b=3

den301095 [7]

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4 years ago

A polynomial subtracted from a polynomial is a polynomial

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5 0

4 years ago

Find a solution to y'(t) = te^-t satisfying the condition y(1) = 1.

Alex_Xolod [135]

Answer:

$y=-e^{-t}(t+1)+1+\frac{2}{e}$

Step-by-step explanation:

The given differential equation is

$y'(t)=te^{-t}$

It can be written as

$\frac{dy}{dt}=te^{-t}$

$dy=te^{-t}dt$

Integrate both sides.

$\int dy=\int te^{-t}dt$

Apply ILATE rule on right side. Here, t is first function and $e^{-t}$ is the second function.

$y=t\int e^{-t}-\int (\frac{d}{dt}t\int e^{-t})$

$y=-te^{-t}-\int (1\times (-e^{-t}))$ $\int e^{-x}=-e^{-x}+C$

$y=-te^{-t}+\int e^{-t}$

$y=-te^{-t}-e^{-t}+C$ .... (1)

Initial condition is y(1) = 1. It means at t=1 the value of y is 1.

$1=-(1)e^{-t}-e^{-(1)}+C$

$1=-e^{-1}-e^{-1}+C$

$1=-2e^{-1}+C$

$1=-\frac{2}{e}+C$

Add $\frac{2}{e}$ on both sides.

$1+\frac{2}{e}=C$

Substitute the value of C in equation (1).

$y=-te^{-t}-e^{-t}+1+\frac{2}{e}$

$y=-e^{-t}(t+1)+1+\frac{2}{e}$

Therefore, the solution of given initial value problem is $y=-e^{-t}(t+1)+1+\frac{2}{e}$ .

4 0

3 years ago