A curve passes through the point (7,6) and has the property that the slope of the curve at every point P is 4 times the y-coordinate of P. What is the equation of the curve? Simplify the equation as much as possible.

dy/dx = 4y

separate variables,
dy/y = 4dx
Integrate
log(y)=4x+C
or y=Ce^4x
At x=7, y=6, so
6=Ce^(4*7)
C=6/e^28
So
y=6e^(4x-28)
Check:
y'(x)=dy/dx=24e^(4x-28)
y'(2.5)=3.655*10^-7
y(2.5)=9.138*10^-8
y'(2.5)/y(2.5)=4.000 ok

To find the equation of the curve, we need to consider the slope of the curve at every point. We are given that the slope of the curve at any point P is 4 times the y-coordinate of P.

Let's denote the x-coordinate of a generic point on the curve as x and the y-coordinate as y. We know that the slope at this point is 4y.

We can use this information to find the equation of the curve using calculus. The derivative of y with respect to x gives us the slope at any point on the curve.

Taking the derivative of y with respect to x, we have:
dy/dx = 4y

This is a separable differential equation. We can rewrite it as:
dy/y = 4dx

Now, let's integrate both sides of the equation. Integrating the left side gives us the natural logarithm of y, and integrating the right side gives us 4x plus a constant of integration, C:
ln(y) = 4x + C

To eliminate the logarithm, we can exponentiate both sides using the property: e^(ln(y)) = y. This gives us:
y = e^(4x+C)

Since e^C is just a constant, let's rewrite it as another constant, A:
y = Ae^(4x)

Now, we can use the given point (7,6) to find the value of A. Plugging in x=7 and y=6 into the equation, we have:
6 = Ae^(4*7)

Simplifying the exponent, we get:
6 = Ae^28

To solve for A, divide both sides of the equation by e^28:
A = 6 / e^28

Finally, we can substitute the value of A back into the equation of the curve:
y = (6 / e^28) * e^(4x)

Simplifying further, we have:
y = 6e^(4x - 28)

Therefore, the equation of the curve passing through the point (7,6) is: y = 6e^(4x - 28).