Solve the initial value problem.?

Question

dy/dx = 6x^3, y(0) = 8

Answer 1

dy/dx=6x^3, y(0)=8

=>

dy=6(x^3)dx

=>

y=6S(x^3)dx+C

=>

y=6[(x^4)/4]+C

=>

y=3(x^4)/2+C

is the G.S.

8=3(0)/2+C

=>

C=8

=>

y=3(x^4)/2+8

is the particular solution.

Answer 2

dy/dx = 6x³

y = 6.(1/4).x⁴ + k → where k in constant

y = (3/2).x⁴ + k → given that: y(0) = 8

(3/2).0⁴ + k = 8

k = 89

→ y = (3/2).x⁴ + 8

Answer 3

Give each problem a try before posting.

If stuck at least try to learn the method for next time

y = (3/2)x^4 + C

when x = 0 y = 8, so C = 8

y = (3/2)x^4 + 8

Answer 4

You already asked many times such questions and were fully answered to you.

Looks like that you are not trying at all. Is this the case? Are you just lazy or trolling?

dy/dx = 6x^3

Integrate both sides. What do you get?

Answer that & we will take it from there. 

Hopefully no one will spoil you the answer. That would be very irresponsible of them. 

Answer 5

dy/dx = 6x^3 ---> dy = 6x^3 dx --->  y = 6/4  x^4 + C   apply IC

y(0) = 8 --->  C = 8 --->   y = 3/2 x^4 + 8

Answer 6

Separate variables:

dy = 6x^3

Integrate:

y = (3/2)(x^4) + C

Plug in initial condition:

8 = C

y = (3/2)(x^4) + 8

Answer 7

dy/dx = 6x³

∫dy = ∫ 6x³ dx

y(x) = 6x⁴/4 + C

y(x) = (3/2)x⁴ + C ......(1)

Given y(0) = 8

y(0) = (3/2)(0)⁴ + C

8 = C

y(x) = (3/2)x⁴ + 8