Math: Antiderivative!! totally lost!! need help!?

Question

dy/dx = 10 - 8x; y(0) = 40

What is the particular antiderivative of each derivative that satisfies the given condition???

Answer 1

I don't know why you are lost. :( Perhaps because there is only a single antiderivitive that satisfies the given conditions??

dy/dx = 10 -8x has lots of solutions,

when you add y(0) = 40 then all but one are eliminated.

Answer 2

ok so you know how to do a basic antiderivative right?

in short, you find out the derivate of what function, y(x), equals y'(x)

so for example, the antiderivate of 2x is x^2, because if you take the deriv of x^2 you get x.

Anyways, your problem is giving you the derivative (10-8x) and you're supposed to find out when the antiderivative equals 40 when you plug in 0.


So first things first is finding the antideriv of y'(10-8x), which is y(x)=10x - 4x^2 + C

The importance here is that C. The C is a constant and you can plug in any number for it, since when you take the derivative of a constant it is simply zero.

So, if we plug in 0 for x, y(0), we get y(0)= 10(0) -4(0)^2 + C

Which equals y(0)= 0+0+C

Since we know that y(0) has to equal 40, so 0+0+C=40,

and therefore C=40.

So plug that into the original y(x) equation, and you come up with f(x)=10x - 4x^2 + 40


Hope this helps!

Answer 3

Integrating dy/dx =10 - 8x gives
y = 10x -4x² + C.
Now use your initial condition to get C.
Plug in your point (0,40):
40 = 0 + 0 + C.
So C= 40 and the answer is
y = 10x-4x²+40.

Answer 4

dy/dx = 10 - 8x

y = 10x - 4x² + C

y(0) = 40, i.e., when x = 0, y = 40

Therefore

y = 10x - 4x² + 40