I tried substituting the $4$ into the $3x-5$ equation, so my slope would be represented as $3(4)-5 = 7$. Then my equation for the line would be $y-3 = 7(x-4)$. That means the equation of the line would be $y = 7x - 25$. However, I'm trying to submit this answer online and it comes back as incorrect. Any help would be appreciated, thank you.
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$\begingroup$You have the slope is $7$ at the point $(4,3)$ which is correct. But, what you found was the equation of the tangent line to the curve at the point $(4,3)$.
SOLUTION: We need to integrate $\frac{dy}{dx}$ to get the function of the original curve. So, $$y=\int (3x-5) \ dx=\frac{3}{2}x^2-5x+C.$$
Now, we plug in the point $(4,3)$ to find $C$. $$3=\frac{3}{2}(4^2)-5(4)+C.$$
So, $C=-1$ and $$y=\frac{3}{2}x^2-5x-1.$$
$\endgroup$ $\begingroup$Your equation only satisfies $y'(x) = 3x-5$ at the point $(4,3)$, rather than everywhere. You need to take the antiderivative to get the family of curves $y = (3/2)x^2 - 5x + C$ that satisfy the given differential equation everywhere. Solve for $C$ by plugging in $(4,3)$ to this family, instead.
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