DY/DX maths question?

Question

Given that

dy/dx = 3√x - x^2

and that y=2/3 when x = 1 find the value of y when x =4

thanks

Answer 1

If dy/dx = 3√x - x^2, then y = 2x^(3/2) - (1/3)x^3 + c by integration.

To find the constant c, plug in the value of x and y at the given point, (1, 2/3) and solve. 2/3 = 2 - 1/3 + c or c = -1.

So your equation is y = 2x^(3/2) - (1/3)x^3 - 1. To find the value of y when x =4, you plug in 4 for x and solve.

y = 2(4)^(3/2) - (1/3)(4)^3 -1
= ?

Good luck!

Answer 2

If dy/dx = 3x^(1/2) - x^2, then multiply both sides by dx and integrate. I get y = 2x^(3/2) - (1/3)x^3 + C. We can solve for C using the fact that y = 2/3 when x = 1, which tells us 2/3 = 2*1^(3/2) - (1/3)*1^3 + C ==> 2/3 = 2 - 1/3 + C = => 2/3 = 5/3 + C ==> C = -3/3 = -1. So when x = 4, y = 2*4^(3/2) - (1/3)*4^3 - 1 = 2*8 - (1/3)*64 = 16 - 64/3 = -16/3.

If the original equation was actually dy/dx = 3*sqrt(x - x^2), it's a harder integration.

Answer 3

If dy/dx=3√x-x^2
then dy=(3√x-x^2)dx
integrate it from 1 to x to find y as a function of x
y(x)=sum(1,x)((3√x-x^2)dx)
y(x)=2x^(3/2)-1/3x^3 - y(1)
y(x)=2x^(3/2)-x^3/3-2/3
y(x)=2x√x-x^3/3-2/3

Therefore:
y(4)=2*4*2-64/3-2/3
y(4)=48/3-64/3-2/3
y(4)=-18/3
y(4)=-6

Answer 4

dy/dx = 3√x - x^2

y = ∫3x^(½ ) - x² dx

= 3x^(3/2) / 3/2 - x³/3 + c

= 2x^(3/2) - x³/3 + c

y = 2/3 when x = 1

2/3 = 2 - 1/3 + c...................c = -1

y = 2x^(3/2) - x³/3 -1

when x = 4

y = 2 . (8) - 4³/3 -1

=15 - 64/3

= - 19/3

Answer 5

dy/dx = 3√x - x^2
y = integrate (3√x - x^2) w.r.t. x
y = integrate (3x^(1/2) - x^2) w.r.t. x
y = [3x^(3/2)]/(3/2) - (x^3)/3 + C
y = 2x^(3/2) - (x^3)/3 + C

When y=2/3 and x=1,
2/3 = 2 - 1/3 + C
C = -1

So the equation is:
y = 2x^(3/2) - (x^3)/3 - 1

When x=4,
y = 2(4^(3/2)) - (4^3)/3 - 1
y = 2(8) - 64/3 - 1
y = 16 - 21 1/3 - 1
y = -6 1/3

Answer 6

dy/dx = 3.x^(1/2) - x²
y = 2.x^(3/2) - x³ / 3 + C
(2/3) = 2 - 1/3 + C
C = - 1
y = 2.x^(3/2) - x³ / 3 - 1
y(4) = 16 - 64/3 - 1
y(4) = 45/3 - 64/3
y(4) = - 19 / 3

Answer 7

dy/dx = 3√x - x^2
y=2x^3/2-x^3/3.+C
whenx=1,y=2/3, 2/3=2-1+C, so C=-1/3
y=2x^3/2-x^3/3-1/3
when x=4
y=16-64/3-1/3=16-65/3=-17/3answer

Answer 8

dy/dx = 3√x - x^2
==> y = 2(x^ 3/2) - (x^3)/3 + c
if x=1 ==> y=2/3
so:
2/3 = 2 - 1/3 +c ==> c= -1
==> y = 2(x^ 3/2) - (x^3)/3 -1
x=4 ==> y = 2*8 - 64/3 - 1 = -19/3

Answer 9

dy/dx = 3√x - x^2
=> y = 3(x^ 3/2)/3/2 - (x^3)/3 + c
=> y = 2(x^ 3/2) - (x^3)/3 + c

when x=1 => y=2/3

=>2/3 = 2 - 1/3 +c
=>2/3 - 2 + 1/3 = c
=> - 1 = c

=> y = 2(x^ 3/2) - (x^3)/3 -1
when x=4
=>y = 2(4)^(3/2) - (1/3)(4)^3 -1
=> y = 2*8 - 64/3 - 1 = -19/3
=> y = -19/3

Answer 10

the curve c has equation y=2x-8squarerootx+5,x≥0