Find the Particular Solution for ds/dt = 14t^2+3t-3 with s=124 at t=0

[hallie]

Find the particular solution determined by the given condition:

ds/dt = 14t^2+3t -3; s=124 when t = 0

Could someone please point me in the right direction to start this problem? I'm not even sure if I titled this post correctly. :P

Thank you!

 

[topsquark]

Find the particular solution determined by the given condition:

ds/dt = 14t^2+3t -3; s=124 when t = 0

Could someone please point me in the right direction to start this problem? I'm not even sure if I titled this post correctly. :P

Thank you!

Hint: \(\displaystyle \int \frac{ds}{dt}~dt = \int ( 14t^2+3t -3)~dt\)

And the LHS is equal to s(t).

-Dan

 

[MarkFL]

You could also set it up using definite integrals by switching the dummy variables of integration and using the given boundaries as the limits:

\(\displaystyle \int_{124}^{s(t)} \,du=\int_0^t 14v^2+3v -3\,dv\)

 

[soroban]

Find the particular solution determined by the given condition:

$$\frac{ds}{dt} \: =\: 14t^2+3t -3;\;\; s=124 \text{ when }t = 0.$$

Could someone please point me in the right direction to start this problem?

It seems that you are new to this type of problem.
Do you understand the given statements?

We are given the derivative of the s-function
and we are asked to find the s-function.

You should know that we will integrate the given function,

$$\int(14t^2 + 3t - 3)\,dt \;=\;\tfrac{14}{3}t^3 + \tfrac{3}{2}t^2 - 3t + C$$

We must evaluate that constant, using the given initial condition.
$$\text{When }t = 0, s = 124.$$

So we have: $$\tfrac{14}{3}(0^3) + \tfrac{3}{2}(0^2) - 3(0) \:=\:124 \quad\Rightarrow\quad C = 124$$

Answer: .$$s(t) \;=\;\tfrac{14}{3}t^3 + \tfrac{3}{2}t^2 - 3t + 124$$