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3 years ago

Find the exact solution for the following system of equations. 2x+3y=48 x+6y=11

Mathematics

1 answer:

TEA [102]3 years ago

4 0

Question 2x+3y=48
Answer x = 24 + -1.5y

Question x+6y=11
Answer x = 11 + -6y

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3/5 ÷ 7/6

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3 years ago

family of solutions of the second-order DE y y 0. Find a solution of the second-order IVP consisting of this differential equati

Anna35 [415]

Answer:

$y = \frac{1}{2}e^x -\frac{1}{2} e^{2-x}$

Step-by-step explanation:

Given

$y=c_1e^x +c_2e^{-x$

$y(1) = 0$

$y'(1) =e$

Required

The solution

Differentiate $y=c_1e^x +c_2e^{-x$

$y' = c_1e^x - c_2e^{-x}$

Next, we solve for c1 and c2

$y(1) = 0$ implies that; x = 1 and y = 0

So, we have:

$y=c_1e^x +c_2e^{-x$

$0 = c_1 * e^1 + c_2 * e^{-1}$

$0 = c_1 e + \frac{1}{e}c_2$ --- (1)

$y'(1) =e$ implies that: x = 1 and y' = e

So, we have:

$y' = c_1e^x - c_2e^{-x}$

$e = c_1 * e^1 - c_2 * e^{-1}$

$e = c_1 e - \frac{1}{e}c_2$ --- (2)

Add (1) and (2)

$0 + e = c_1e + c_1e + \frac{1}{e}c_2 - \frac{1}{e}c_2$

$e = 2c_1e$

Divide both sided by e

$1 = 2c_1$

Divide both sides by 2

$c_1 = \frac{1}{2}$

Substitute $c_1 = \frac{1}{2}$ in $0 = c_1 e + \frac{1}{e}c_2$

$0 = \frac{1}{2} e+ \frac{1}{e}c_2$

Rewrite as:

$\frac{1}{e}c_2 = -\frac{1}{2} e$

Multiply both sides by e

$c_2 = -\frac{1}{2} e^2$

So, we have:

$y=c_1e^x +c_2e^{-x$

$y = \frac{1}{2}e^x -\frac{1}{2} e^2 * e^{-x}$

$y = \frac{1}{2}e^x -\frac{1}{2} e^{2-x}$

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3 years ago