Solve the initial value problem. $$ \begin{array}{l}\frac{d s}{d t}=(8 t+3) \sin \left(4 t^{2}+3 t ight), s(0)=-8 \ s=-\cos (8 t+3)-\frac{13}{4} \ s=\sin \left(4 t^{2}+3 t ight)-8 \ s=-\cos \left(4 t^{2}+3 t ight)-7 \ s=\cos \left(4 t^{2}+3 t ight)-9\end{array} $$ $$ \begin{array}{l}s \ s,\end{array} $$

Question

Solve the initial value problem. $$ \begin{array}{l}\frac{d s}{d t}=(8 t+3) \sin \left(4 t^{2}+3 tight), s(0)=-8 \ s=-\cos (8 t+3)-\frac{13}{4} \ s=\sin \left(4 t^{2}+3 tight)-8 \ s=-\cos \left(4 t^{2}+3 tight)-7 \ s=\cos \left(4 t^{2}+3 tight)-9\end{array} $$ $$ \begin{array}{l}s \ s,\end{array} $$

UpStudy AI · Accepted Answer

We start with the differential equation
$$
\frac{d s}{d t}=(8t+3)\sin(4t^2+3t)
$$
with initial condition
$$
s(0)=-8.
$$

**Step 1.** Perform a substitution:
Let
$$
u = 4t^2+3t.
$$
Then,
$$
\frac{du}{dt} = 8t+3.
$$
Thus, the differential equation becomes
$$
\frac{ds}{dt} = \sin(u)\frac{du}{dt},
$$
or equivalently,
$$
ds = \sin(u)\,du.
$$

**Step 2.** Integrate both sides with respect to their variables:
$$
\int ds = \int \sin(u)\,du.
$$
Performing the integration gives
$$
s = -\cos(u) + C,
$$
where $$C$$ is the constant of integration.

**Step 3.** Substitute back $$u = 4t^2+3t$$ to express $$s$$ in terms of $$t$$:
$$
s(t) = -\cos(4t^2+3t) + C.
$$

**Step 4.** Apply the initial condition $$s(0)=-8$$:
$$
-8 = -\cos\bigl(4(0)^2+3(0)\bigr) + C = -\cos(0) + C.
$$
Since $$\cos(0)=1$$, we have
$$
-8 = -1 + C.
$$
Solving for $$C$$ gives
$$
C = -8 + 1 = -7.
$$

**Final Answer:**
The solution to the initial value problem is
$$
s(t) = -\cos(4t^2+3t)-7.
$$
The solution to the initial value problem is $$ s(t) = -\cos(4t^2 + 3t) - 7 $$.

Solve the initial value problem.

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