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# Solve the initial value problem.?

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

### 7 Answers

- PinkgreenLv 72 months ago
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.

- la consoleLv 72 months ago
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

- Ian HLv 72 months ago
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

- rotchmLv 72 months ago
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.

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- nyphdinmdLv 72 months ago
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

- stanschimLv 72 months ago
Separate variables:

dy = 6x^3

Integrate:

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

Plug in initial condition:

8 = C

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

- AshLv 72 months ago
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