Intergrate 3X/8 ...D2y/dx2

[dinomath]

Intergrate 3X/8 ...D2y/dx2 DY = 3 & Y=4 WHEN X = 4


I don't know how to intergrate from 3x power over a number ? Is this 3x^2/2 then divide by 8 ?

Also

dy/dx of 2^X

I am really don't know how to solve a power of x over a power of 2

Thank YOU For you help !!!

 

[fasteddie65]

Intergrate 3X/8 ...D2y/dx2 DY = 3 & Y=4 WHEN X = 4


I don't know how to intergrate from 3x power over a number ? Is this 3x^2/2 then divide by 8 ?

Also

dy/dx of 2^X

I am really don't know how to solve a power of x over a power of 2

Thank YOU For you help !!!

I think I understand what your problem is asking, but it is not totally clear.

If you integrate (NOTE SPELLING!) d^y/dx^2 = 3x/8 dx, you get dy/dx = 3x^2/16 + C.
I think you are saying that dy/dx = 4 when x = 4, so 4 = 3(4^2)/16 + C = 3 + C, so C = 1.
That takes us to dy/dx = 3x^2/16 + 1

Integrating again, you get y = x^3/16 + x + C'. 4 = 4^3/16 + 4 + C' = 4 + 4 + C', so C' = -4

The answer I get is y = x^3/16 + x - 4.

 

[Deleted member 4993]

Intergrate 3X/8 ...D2y/dx2 DY = 3 & Y=4 WHEN X = 4


I don't know how to intergrate from 3x power over a number ? Is this 3x^2/2 then divide by 8 ?

Also

dy/dx of 2^X

Let

y = 2[sup:22vmxw0u]x[/sup:22vmxw0u]

ln(y) = x * ln(2)

Now apply implicit differentiation and differentiate both sides....