Question

If xy + 7ey = 7e, find the value of y'' at the point where x = 0.

Answer 1

Answer:

Please read my updated answer.

I am wasn't sure what you were asking so here and 2 answers.

1st answer $y=\frac{1}{e}$

2nd answer $\frac{dy}{dx}=-\frac{y}{x+7e}$

Step-by-step explanation:

$xy+7ey=7$

$when \\ x=0$

Substitute $0$ for $x$ then simplify.

$(0*y)+7ey=7\\ 0+7ey=7\\ 7ey=7$

Divide each term in $7ey=7$ by $7e$ and simplify.

$\frac{7ey}{7e}=\frac{7}{7e}$

Cancel the common factor of $7$ on the left side.

Rewrite the expression.

$\frac{ey}{e}=\frac{7}{7e}$

Cancel the common factor of $e$ on the left side.

$y=\frac{7}{7e}$

Cancel the common factor of 7 on the right side.

Rewrite the expression.

$y=\frac{1}{e}$

2nd Answer

Step-by-step explanation:

$xy+7ey=7$

Differentiate both sides of the equation.

$\frac{d}{dx} (xy+7ey)=\frac{d}{dx}(7e)$

Differentiate the left side of the equation.

By the Sum Rule, the derivative of $xy+7ey$ with respect to x is $\frac{d}{dx} [xy]+\frac{d}{dx}[7ey]$

Evaluate $\frac{d}{dx} [xy]$

Differentiate using the Product Rule

Rewrite $\frac{d}{dx} [xy]$ as $y'$.

$xy'+y+\frac{d}{dx} [7ey]$

Evaluate $\frac{d}{dx} [7ey]$

Since $7e$ is constant with respect to $x$, the derivative of $7e$ with respect to $x$ is $0$.

Reform the equation by setting the left side equal to the right side.

$xy'+7ey'+y=0$

Solve for $y'$.

Subtract $y$ from both sides of the equation.

$xy'+7ey'=-y$

Factor $y'$ out of $xy'$.

$y'x+7ey'=-y$

Factor $y'$ out of $7ey'$.

$y'(x)+y'(7e)=-y$

Factor out $y'$ out of  $y'(x)+y'(7e)$

$y'(x+7e)=-y$

Divide each term in $y'(x+7e)=-y$ by $x+7e$ and simplify.

$\frac{y'(x+7e)}{x+7e} =\frac{-y}{x+7e}$

Cancel the common factor of $x+7e$.

$y'=-\frac{-y}{x+7e}$

Replace $y'$ with  $\frac{dy}{dx}=-\frac{y}{x+7e}$

$\frac{dy}{dx}=-\frac{y}{x+7e}$