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anyanavicka [17]
4 years ago
12

Solve the given system of differential equations by systematic elimination

Mathematics
1 answer:
Korolek [52]4 years ago
3 0

\begin{cases}(D^2+2)x-2y=0\\-2x+(D^2+5)y=0\end{cases}

Solve for y(t) in the first ODE:

y=\dfrac{(D^2+2)x}2

Substitute this into the second ODE:

-2x+(D^2+5)\dfrac{(D^2+2)x}2=0

-4x+(D^4+7D^2+10)x=0

(D^4+7D^2+6)x=0

This ODE is linear with characteristic equation

r^4+7r^2+6=(r^2+6)(r^2+1)=0

with roots at r=\pm i\sqrt6 and r=\pm i, so that the characteristic solution x_c(t) is

\boxed{x(t)=C_1\cos(\sqrt6\,t)+C_2\sin(\sqrt6\,t)+C_3\cos t+C_4\sin t}

and from here we can solve for y(t):

y=\dfrac{(D^2+2)x}2

\boxed{y(t)=-2C_1\cos(\sqrt6\,t)-2C_2\sin(\sqrt6\,t)+\dfrac{C_3}2\cos t+\dfrac{C_4}2\sin t}

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The cost of the brush is $2.25, the cost of the sketchbook is $13.50, and the cost of the paint set is $18.

<h3>What is a linear equation?</h3>

It is defined as the relation between two variables, if we plot the graph of the linear equation we will get a straight line.

If in the linear equation, one variable is present, then the equation is known as the linear equation in one variable.

Let's suppose the brush cost is $B, the sketchbook cost is $S, and the paint set cost is $P.

From the problem, we can draw three linear equations in three variables:

The brush was 1/6 as much as the sketchbook:

B = (1/6)S or

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S = (3/4)P  or

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The total cost would be:

B+S+P = 35.75 - 2

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After solving equations (1), (2), and (3) we will get the cost of each item:

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Thus, the cost of the brush is $2.25, the cost of the sketchbook is $13.50, and the cost of the paint set is $18.

Learn more about the linear equation here:

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Simple.....

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For a party, you can spend no more than $20 on cakes. Egg cake costs $4 and cream cake costs $2. Write the linear inequality tha
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Answer:

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