# Solve the initial value problem.?

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

Relevance
• 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.

• 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

• 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

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.

• 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

• Separate variables:

dy = 6x^3

Integrate:

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

Plug in initial condition:

8 = C

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

• 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