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Thank you Alan,  I shall study your solution,   

Here is my solution:

Solve:

\(y''-2y'+y=(x^2+1)e^x\)

Big Hint:

The solution will be of the from  

\(y=(Ax^4+Bx^3+Cx^2)e^x \\~\\ \text{Where A,B and C are constants.}\)

\( y=(Ax^4+Bx^3+Cx^2)e^x\\~\\ y'=[Ax^4+(4A+B)x^3+(3B+C)x^2+2Cx]e^x\\~\\ y''=[Ax^4+(4A+4A+B)x^3+(12A+3B+3B+C)x^2+(6B+2C+2C)x+2C]e^x\\ y''=[Ax^4+(8A+B)x^3+(12A+6B+C)x^2+(6B+4C)x+2C]e^x\\~\\ y''-2y'+y\\ \begin{array}{rrrrrrr} =e^x[&Ax^4&+&(8A+B)x^3&+&(12A+6B+C)x^2&+&(6B+4C)x&+&2C&\\ &+(-2A)x^4&+&(-8A-2B)x^3&+&(-6B-2C)x^2&+&(-4C)x&\\ &+Ax^4&+&Bx^3&+&Cx^2&&&&&]\\~\\ =e^x[&0x^4&+&0x^3&+&12Ax^2&+&6Bx&+&2C&]\\~\\ \end{array}\\ so\\ y''-2y'+y=e^x[12Ax^2+6Bx+2C]\\~\\ \)

\([12Ax^2+6Bx+2C]e^x=(x^2+1)e^x\\~\\ 12A=1 \qquad 6B=0 \qquad 2C=1\\ A=\frac{1 }{12}\qquad \;\; B=0 \qquad \;\; C=\frac{1}{2}\\~\\ so\;\;if\;\;\\ y''-2y'+y=(x^2+1)e^x\\ then\\ y=\left(\dfrac{x^4}{12}+\dfrac{x^2}{2}\right)e^x \)

Like so:

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Here's my worked solution:

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what's sin when tan = -2?

\(\large sinx=\pm\frac{tanx}{\sqrt{1+tan^2x}}\)

\(sinx=\pm\frac{-2}{\sqrt{1+(-2)^2}}\)

\(sinx=\frac{\pm2}{\sqrt5}=\pm0.894427191\)

\(When \ tanx =-2 \ \ then \ \ sinx=\pm0.894427191.\)

  !

1000

sqrt(5x-4)-sqrt(x)=2

\(\begin{array}{|rcll|} \hline \sqrt{5x-4}-\sqrt{x} &=& 2 \quad &| \quad +\sqrt{x} \\ \sqrt{5x-4} &=& 2 +\sqrt{x} \quad &| \quad \text{square both sides} \\ 5x-4 &=& (2 +\sqrt{x})^2 \\ 5x-4 &=& 4+4\cdot \sqrt{x} + x \quad &| \quad -x \\ 4x-4 &=& 4+4\cdot \sqrt{x} \quad &| \quad :4 \\ x-1 &=& 1+ \sqrt{x} \quad &| \quad -1\\ x-2 &=& \sqrt{x} \quad &| \quad \text{square both sides} \\ (x-2)^2 &=& x\\ x^2-4x+4 &=& x \quad &| \quad -x \\ x^2-5x+4 &=& 0\\\\ x &=& \frac{5\pm \sqrt{25-4\cdot 4} } {2} \\ x &=& \frac{5\pm \sqrt{25-16} } {2} \\ x &=& \frac{5\pm \sqrt{9} } {2} \\ x &=& \frac{5\pm 3 } {2} \\\\ x_1 &=& \frac{5 + 3 } {2} \\ x_1 &=& \frac82 \\ \mathbf{x_1} & \mathbf{=} & \mathbf{4} \\\\ x_2 &=& \frac{5 - 3 } {2} \\ x_2 &=& \frac22 \\ \mathbf{x_2} & \mathbf{=} & \mathbf{1} \quad \text{ no solution}\\ \hline \end{array}\)

So, you have 2 2/3. And 3/3 make a whole. If you have two wholes, it's like two 3/3. So you'll multiply 2 by 3, which is six, so at this point you have 6/3 and 2/3. Add those together and now you have 8/3. Hope this helps!

oh, nuts, here we go again with numbers and letters............ are the exponets multiplied or is there exponets of exponets here?

Actually, I simplified it (if you can call this long number simplifying) so here it is: (6658042.581874427*x)*y

It's x^(x62y - 1) • x62y • 2^(x62y + 1)/3

At least, that is the derivative. There's no more simplifying that can be done.

Yes it is theoretical in that friction is ignored.....but GRAVITY is used in the equation  ( the 'g'  in 'mgh' )  

The 16 ounce bottle, because it is 19.9 cents for each ounce in the super-cleaner, but it is 20.6 cents for the wonder cleaner.

Power is  Joules per second (watts)

Power = j/s = 50/100,000 = .0005 watts    (not much)

68+1=69

6284562938476592837465897346598036459078345908273489572.

Divide dollars by ounces in both cases and compare the unit price.

Never mind, I'm wrong, he's right.

then there would be 3,500 men and woman, .7 of them are women, so 3500 * .7 = 2500, (approximately, you can't have 0.000000005 a man) so 3500-2500 (2500 is the woman) = 1000, so there are 1000 men

4,000 - 500 =3,500 total of men and women

7/[5+7] x 3,500 =2,042 number of women!!!

5/[5+7] x 3,500 =1,458 number of men!!!