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hodyreva [135]
3 years ago
7

5y(4y^2−2y)+3y(−2y^2+4y)Please help

Mathematics
1 answer:
Alona [7]3 years ago
5 0

Answer:

factor the expression =2y^2 (7y+1) or simplify it=14y^3 + 2y^2

Step-by-step explanation:

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X~4y=11 , 5x~7y=~10 system of equation using elimination to solve
tino4ka555 [31]
1x + 4y = 11 ⇒ 5x + 20y = 55
5x - 7y = -10 ⇒ <u>5x -   7y = -10</u>
                                 <u>27y</u> = <u>65</u>
                                  27     27
                                     y = 2¹¹/₂₇ 
                     x + 4(2¹¹/₂₇) = 11
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                         <u>    - 8²²/₂₇  - 8²²/₂₇
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3 0
3 years ago
Use the method of reduction of order to find a second independent solution of the given differential equation. t2y'' + 3ty' + y
Charra [1.4K]

I assume you mean y_1=t^{-1}, and not y_1=t-1, since this doesn't satisfy the ODE.

Assume a second solution of the form y_2=vy_1, where v is a function of t. Then

{y_2}'=v'y_1+v{y_1}'

{y_2}''=v''y_1+2v'{y_1}'+v{y_1}''

Substituting into the ODE gives

t^2\left(\dfrac{v''}t-\dfrac{2v'}{t^2}+\dfrac{2v}{t^3}\right)+3t\left(\dfrac{v'}t-\dfrac v{t^2}\right)+\dfrac vt=0

\implies tv''+v'=0

\implies(tv')'=0

\implies tv'=C

\implies v'=\dfrac Ct

\implies v=C_1\ln|t|+C_2

\implies y_2=\dfrac{\ln t}t

where we omit the second term because it's already accounted for by y_1.

3 0
3 years ago
Solve the system of equations by row-reduction. At each step, show clearly the symbol of the linear combinations that allow you
adell [148]

Answer:

1) The solution of the system is

\left\begin{array}{ccc}x_1&=&5\\x_2&=&8\\x_3&=&-13\end{array}\right

2) The solution of the system is

\left\begin{array}{ccc}x_1&=&2\\x_2&=&-7\\x_3&=&-1\end{array}\right

Step-by-step explanation:

1) To solve the system of equations

\left\begin{array}{ccccccc}&3x_2&-5x_3&=&89\\6x_1&&+x_3&=&17\\x_1&-x_2&+8x_3&=&-107\end{array}\right

using the row reduction method you must:

Step 1: Write the augmented matrix of the system

\left[ \begin{array}{ccc|c} 0 & 3 & -5 & 89 \\\\ 6 & 0 & 1 & 17 \\\\ 1 & -1 & 8 & -107 \end{array} \right]

Step 2: Swap rows 1 and 2

\left[ \begin{array}{ccc|c} 6 & 0 & 1 & 17 \\\\ 0 & 3 & -5 & 89 \\\\ 1 & -1 & 8 & -107 \end{array} \right]

Step 3:  \left(R_1=\frac{R_1}{6}\right)

\left[ \begin{array}{ccc|c} 1 & 0 & \frac{1}{6} & \frac{17}{6} \\\\ 0 & 3 & -5 & 89 \\\\ 1 & -1 & 8 & -107 \end{array} \right]

Step 4: \left(R_3=R_3-R_1\right)

\left[ \begin{array}{ccc|c} 1 & 0 & \frac{1}{6} & \frac{17}{6} \\\\ 0 & 3 & -5 & 89 \\\\ 0 & -1 & \frac{47}{6} & - \frac{659}{6} \end{array} \right]

Step 5: \left(R_2=\frac{R_2}{3}\right)

\left[ \begin{array}{ccc|c} 1 & 0 & \frac{1}{6} & \frac{17}{6} \\\\ 0 & 1 & - \frac{5}{3} & \frac{89}{3} \\\\ 0 & -1 & \frac{47}{6} & - \frac{659}{6} \end{array} \right]

Step 6: \left(R_3=R_3+R_2\right)

\left[ \begin{array}{ccc|c} 1 & 0 & \frac{1}{6} & \frac{17}{6} \\\\ 0 & 1 & - \frac{5}{3} & \frac{89}{3} \\\\ 0 & 0 & \frac{37}{6} & - \frac{481}{6} \end{array} \right]

Step 7: \left(R_3=\left(\frac{6}{37}\right)R_3\right)

\left[ \begin{array}{ccc|c} 1 & 0 & \frac{1}{6} & \frac{17}{6} \\\\ 0 & 1 & - \frac{5}{3} & \frac{89}{3} \\\\ 0 & 0 & 1 & -13 \end{array} \right]

Step 8: \left(R_1=R_1-\left(\frac{1}{6}\right)R_3\right)

\left[ \begin{array}{ccc|c} 1 & 0 & 0 & 5 \\\\ 0 & 1 & - \frac{5}{3} & \frac{89}{3} \\\\ 0 & 0 & 1 & -13 \end{array} \right]

Step 9: \left(R_2=R_2+\left(\frac{5}{3}\right)R_3\right)

\left[ \begin{array}{ccc|c} 1 & 0 & 0 & 5 \\\\ 0 & 1 & 0 & 8 \\\\ 0 & 0 & 1 & -13 \end{array} \right]

Step 10: Rewrite the system using the row reduced matrix:

\left[ \begin{array}{ccc|c} 1 & 0 & 0 & 5 \\\\ 0 & 1 & 0 & 8 \\\\ 0 & 0 & 1 & -13 \end{array} \right] \rightarrow \left\begin{array}{ccc}x_1&=&5\\x_2&=&8\\x_3&=&-13\end{array}\right

2) To solve the system of equations

\left\begin{array}{ccccccc}4x_1&-x_2&+3x_3&=&12\\2x_1&&+9x_3&=&-5\\x_1&+4x_2&+6x_3&=&-32\end{array}\right

using the row reduction method you must:

Step 1:

\left[ \begin{array}{ccc|c} 4 & -1 & 3 & 12 \\\\ 2 & 0 & 9 & -5 \\\\ 1 & 4 & 6 & -32 \end{array} \right]

Step 2: \left(R_1=\frac{R_1}{4}\right)

\left[ \begin{array}{ccc|c} 1 & - \frac{1}{4} & \frac{3}{4} & 3 \\\\ 2 & 0 & 9 & -5 \\\\ 1 & 4 & 6 & -32 \end{array} \right]

Step 3: \left(R_2=R_2-\left(2\right)R_1\right)

\left[ \begin{array}{ccc|c} 1 & - \frac{1}{4} & \frac{3}{4} & 3 \\\\ 0 & \frac{1}{2} & \frac{15}{2} & -11 \\\\ 1 & 4 & 6 & -32 \end{array} \right]

Step 4: \left(R_3=R_3-R_1\right)

\left[ \begin{array}{ccc|c} 1 & - \frac{1}{4} & \frac{3}{4} & 3 \\\\ 0 & \frac{1}{2} & \frac{15}{2} & -11 \\\\ 0 & \frac{17}{4} & \frac{21}{4} & -35 \end{array} \right]

Step 5: \left(R_2=\left(2\right)R_2\right)

\left[ \begin{array}{ccc|c} 1 & - \frac{1}{4} & \frac{3}{4} & 3 \\\\ 0 & 1 & 15 & -22 \\\\ 0 & \frac{17}{4} & \frac{21}{4} & -35 \end{array} \right]

Step 6: \left(R_1=R_1+\left(\frac{1}{4}\right)R_2\right)

\left[ \begin{array}{cccc} 1 & 0 & \frac{9}{2} & - \frac{5}{2} \\\\ 0 & 1 & 15 & -22 \\\\ 0 & \frac{17}{4} & \frac{21}{4} & -35 \end{array} \right]

Step 7: \left(R_3=R_3-\left(\frac{17}{4}\right)R_2\right)

\left[ \begin{array}{ccc|c} 1 & 0 & \frac{9}{2} & - \frac{5}{2} \\\\ 0 & 1 & 15 & -22 \\\\ 0 & 0 & - \frac{117}{2} & \frac{117}{2} \end{array} \right]

Step 8: \left(R_3=\left(- \frac{2}{117}\right)R_3\right)

\left[ \begin{array}{cccc} 1 & 0 & \frac{9}{2} & - \frac{5}{2} \\\\ 0 & 1 & 15 & -22 \\\\ 0 & 0 & 1 & -1 \end{array} \right]

Step 9: \left(R_1=R_1-\left(\frac{9}{2}\right)R_3\right)

\left[ \begin{array}{cccc} 1 & 0 & 0 & 2 \\\\ 0 & 1 & 15 & -22 \\\\ 0 & 0 & 1 & -1 \end{array} \right]

Step 10: \left(R_2=R_2-\left(15\right)R_3\right)

\left[ \begin{array}{cccc} 1 & 0 & 0 & 2 \\\\ 0 & 1 & 0 & -7 \\\\ 0 & 0 & 1 & -1 \end{array} \right]

Step 11:

\left[ \begin{array}{ccc|c} 1 & 0 & 0 & 2 \\\\ 0 & 1 & 0 & -7 \\\\ 0 & 0 & 1 & -1 \end{array} \right]\rightarrow \left\begin{array}{ccc}x_1&=&2\\x_2&=&-7\\x_3&=&-1\end{array}\right

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