Calculus 1?

Question

Find g, if it is known that g''(q)=1+3sinq, and that g(0)=1 and g'(0)=0

I don't get this question..
solving it for hours
please help :))

q=Ө q=Ө

Answer 1

Find g, if it is known that g''(q)=1+3sinq, and that g(0)=1 and g'(0)=0

You are being asked to find the function g that satisfies all the given conditions (this is sometimes referred to as an initial value problem). What needs to be done first is to work back to g'(q) and then finally to g(q).
To work back to g'(q) we need to antidifferentiate or find the function whose derivative is the function in the original problem.
g'(q)= q-3cos(q) +C where C is some constant. The value of C can be determined using the fact that you know that g'(0)=0.

Substituting 0 in to g'(q) we have g'(0)= 0-3cos(0) +C=0
Solving for C we have 0-3 +C=0 or C=3, thus giving us g'(q)= q-3cos(q)+3

Proceeding in a like manner we need to now find g(q).

g(q)=(q^2)/2-3sin(q)+3q+c where c is some constant (note the different use of variable as the two constants are most likely NOT the same value).

Substituting 0 in to g(q) we have
g(0)=(0^2)/2-3sin(0)+3(0)+c=1.
Solving for c we have 0-0+0+c=1, thus giving us the function g(q) that satisfies all the given conditions

g(q)=(q^2)/2-3sin(q)+3q+1

Answer 2

Integrate g''(q):
g'(q) = q - 3cosq + C
Plug in initial condition to find C:
g'(0) = 0 - 3cos(0) + C = 0
C = 3cos(0)
C = 3

g'(q) = q - 3cosq + 3

Integrate g'(q):
g(q) = (q^2)/2 - 3sinq + 3q + D
Plug in initial condition to find D:
g(0) = (0^2)/2 - 3sin(0) + 3(0) + D = 1
D = 1

So
g(q) = (q^2)/2 - 3sinq + 3q + 1

Answer 3

you must integrate, but leave the "+ c" in each time

g''(q)=1+3sinq

so g'(q)=q-3cos(q)+c

you know g'(0)=0, so plug that into your equation

0=0-3cos(0)+c

when you solve you find c=3

g'(q)=q-3cos(q)+3

integrate again to find g(q)

g(q)=q^2-3sin(q)+3q+c

plug in again g(0)=1

1=0^2-3sin(0)+3(0)+c

so c=1

and there you have

g(q)=q^2-3sin(q)+3q+1

QED

Answer 4

First step:
Calculate the integral of
1+3sinq

Second step:
Calculate integral of that integral.

Third step is easy.