Solve for x: -10x/2+x=x+8/x

1 answer:

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$The\ domain:\\D:\ x\neq-2\ \wedge\ x\neq0\\\\\dfrac{-10x}{2+x}=\dfrac{x+8}{x}\ \ \ \ |\text{cross multiply}\\\\(-10x)(x)=(2+x)(x+8)\\\\-10x^2=(2)(x)+(2)(8)+(x)(x)+(x)(8)\\\\-10x^2=2x+16+x^2+8x\\\\-10x^2=x^2+10x+16\ \ \ +10x^2\\\\11x^2+10x+16=0\\\\a=11,\ b=10,\ c=16\\\\b^2-4ac=10^2-4(11)(16)=100-704=-604 < 0\\\\NO\ REAL\ SOLUTION$

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$y = e^x\\\\\displaystyle y = \sum_{k=1}^{\infty}\frac{x^k}{k!}\\\\\displaystyle y= 1+x+\frac{x^2}{2!} + \frac{x^3}{3!}+\ldots\\\\\displaystyle y' = \frac{d}{dx}\left( 1+x+\frac{x^2}{2!} + \frac{x^3}{3!}+\frac{x^4}{4!}+\ldots\right)\\\\$

$\displaystyle y' = \frac{d}{dx}\left(1\right)+\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left(\frac{x^2}{2!}\right) + \frac{d}{dx}\left(\frac{x^3}{3!}\right) + \frac{d}{dx}\left(\frac{x^4}{4!}\right)+\ldots\\\\\displaystyle y' = 0+1+\frac{2x^1}{2*1} + \frac{3x^2}{3*2!} + \frac{4x^3}{4*3!}+\ldots\\\\\displaystyle y' = 1 + x + \frac{x^2}{2!}+ \frac{x^3}{3!}+\ldots\\\\\displaystyle y' = \sum_{k=1}^{\infty}\frac{x^k}{k!}\\\\\displaystyle y' = e^{x}\\\\$

This shows that $y' = y$ is true when $y = e^x$

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Note 1: A more general solution is $y = Ce^x$ for some constant C.
Note 2: It might be tempting to say the general solution is $y = e^x+C$ , but that is not the case because $y = e^x+C \to y' = e^x+0 = e^x$ and we can see that $y' = y$ would only be true for C = 0, so that is why $y = e^x+C$ does not work.