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enyata [817]
3 years ago
13

54 can be divided with no remainder by which of the following numbers 2 3 4 5 6 9 10​

Mathematics
1 answer:
Colt1911 [192]3 years ago
8 0

Answer:

6 and 9

Step-by-step explanation:

6 x 9 = 54

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The answer is in the photo
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Mr. Donovan bought a shirt for $18.47<br> and pants for $25.95. How much<br> money did he spend?
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Answer:

$34.42

Step-by-step explanation:

18.47

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What is the mean absolute deviation for 7, 5, 1, 6, 2, and 3?
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3 years ago
Observe that x and e^x are solutions to the homogeneous equation associated with:
yanalaym [24]

To take advantage of the characteristic solutions y_1(x)=x and y_2(x)=e^x, you can try the method of variation of parameters, where we look for a solution of the form

y=y_1u_1+y_2u_2

with the condition that

{u_1}'y_1+{u_2}'y_2=0

\implies{u_1}'x+{u_2}'e^x=0 (\mathbf 1)

Then

y'={y_1}'u_1+y_1{u_1}'+{y_2}'u_2+y_2{u_2}'

\implies y'={y_1}'u_1+{y_2}'u_2

y''={y_1}''u_1+{y_1}'{u_1}'+{y_2}''u_2+{y_2}'{u_2}'

Substituting into the ODE gives

(1-x)({y_1}''u_1+{y_1}'{u_1}'+{y_2}''u_2+{y_2}'{u_2}')+x({y_1}'u_1+{y_2}'u_2)-y_1u_1+y_2u_2=2(x-1)^2e^{-x}

Since

y_1=x\implies{y_1}'=1\implies{y_1}''=0

y_2=e^x\implies{y_2}'=e^x\implies{y_2}''=e^x

the above reduces to

(1-x)({u_1}'+e^x{u_2}')=2(x-1)^2e^{-x}

{u_1}'+e^x{u_2}'=2(1-x)e^{-x} (\mathbf 2)

(\mathbf 1) and (\mathbf 2) form a linear system that we can solve for {u_1}',{u_2}' using Cramer's rule:

{u_1}'=\dfrac{W_1(x)}{W(x)},{u_2}'=\dfrac{W_2(x)}{W(x)}

where W(x) is the Wronskian determinant of the fundamental system and W_i(x) is the same determinant, but with the i-th column replaced with (0,2(x-1)^2e^{-x}).

W(x)=\begin{vmatrix}x&e^x\\1&e^x\end{vmatrix}=e^x(x-1)

W_1(x)=\begin{vmatrix}0&e^x\\2(x-1)^2e^{-x}&e^x\end{vmatrix}=-2(x-1)^2

W_2(x)=\begin{vmatrix}x&0\\e^x&2(x-1)^2e^{-x}\end{vmatrix}=2xe^{-x}(x-1)^2

So we have

{u_1}'=\dfrac{-2(x-1)^2}{e^x(x-1)}\implies u_1=2xe^{-x}

{u_2}'=\dfrac{2xe^{-x}(x-1)^2}{e^x(x-1)}\implies u_2=-x^2e^{-2x}

Then the particular solution is

y_p=2x^2e^{-x}-x^2e^{-x}=x^2e^{-x}

giving the general solution to the ODE,

\boxed{y(x)=C_1x+C_2e^x+x^2e^{-x}}

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