I think that jeremy should learn how to do his own math
C: none of these are solutions to the given equation.
• If<em> y(x)</em> = <em>e</em>², then <em>y</em> is constant and <em>y'</em> = 0. Then <em>y'</em> - <em>y</em> = -<em>e</em>² ≠ 0.
• If <em>y(x)</em> = <em>x</em>, then <em>y'</em> = 1, but <em>y'</em> - <em>y</em> = 1 - <em>x</em> ≠ 0.
The actual solution is easy to find, since this equation is separable.
<em>y'</em> - <em>y</em> = 0
d<em>y</em>/d<em>x</em> = <em>y</em>
d<em>y</em>/<em>y</em> = d<em>x</em>
∫ d<em>y</em>/<em>y</em> = ∫ d<em>x</em>
ln|<em>y</em>| = <em>x</em> + <em>C</em>
<em>y</em> = exp(<em>x</em> + <em>C </em>)
<em>y</em> = <em>C</em> exp(<em>x</em>) = <em>C</em> <em>eˣ</em>
Answer:
2) 4,2
Step-by-step explanation:
<u>Factor by grouping:</u>
Setting the expression equal to 0 and using the Zero Product Property, 
and 
, which are the factors of the original polynomial.
