Find the particular solution that satifies the differential equation and the initial condition. f"(x) = x2 f'(0) = 8, f(0) = 4

f(x) = x^4/12 + 8x + 4

Step-by-step explanation:

f"(x) = x^2

Integrate both sides with respect to x

f'(x) = ∫ x^2 dx

= (x^2+1)/2+1

= (x^3)/3 + C

f(0) = 8

Put X = 0

f'(0) = 0+ C

8 = 0 + C

C= 8

f'(x) = x^3/3 + 8

Integrate f(x) again with respect to x

f(x) = ∫ (x^3 / 3 ) +8 dx

= ∫ x^3 / 3 dx + ∫8dx

= x^(3+1) / 3(3+1) + 8x + D

= x^4/12 + 8x + D

f(0) = 4

Put X = 0

f(0) = 0 + 0 + D

4 = D

Therefore

f(x) = x^4 /12 + 8x + 4

step-by-step explanation:

i am not sure sorry : c

option 2 is correct.

step-by-step explanation:

we are given two inequality equation

x+y> 5 and -3x+2y< -2

first we draw the graph of both function using t-table method

equation: x+y> 5

for x=0, y=5

for x=-5, y=0

table:

x   :   0     5

y   :   5     0

test point (0,0) , 0> 5

false

equation: -3x+2y< -2

for x=0, y=-1

for x=2, y=4

table:

x   :   0     2

y   : -1       4

test point (0,0) , 0< -2

true

now we draw the graph. see the attachment for graph.

common region would be solution set.

option 2 is correct.