MHB -t1.13 IVP \frac{ds}{dt}&=12t(3t^2-1)^3 ,s(1)=3\

[karush]

$\tiny{2214.t1.13}$
1000
$\textsf{solve the initial value problem}$
\begin{align*}\displaystyle
\frac{ds}{dt}&=12t(3t^2-1)^3 ,s(1)=3\\
I_{13}&=\int 12t(3t^2-t)^3 \, dt\\
u&=(3t^2-1) \therefore \frac{1}{6t}du=dt \\
&=2\int u^3 du\\
&=2\left[\frac{u^4}{4}\right]\\
\textit{Back Substitute U}&\\
&=2\left[\frac{(3t^2-1)^4}{4}\right]+C\\
\textit{Solve for C}&\\
s(1)&=\left[\frac{(3(1)^2-1)^4}{2}\right]+C=3\\
&=\left[\frac{16}{2}\right]+C=3\\
&=8+C=3 \therefore C=-5\\
\textit{Initial Value}&\\
s&=\color{red}{\frac{1}{2}(3t^2-1)^4-5}
\end{align*}ok took me 2 hours to do this and hope it is ok
I was curious if there is a way to right justify the text inside an align

also suggestions if any

much mahalo ahead

 

[MarkFL]

I moved this here since IVP's are a topic from the study of differential equations.

We are given:

$$\d{s}{t}=12t\left(3t^2-1\right)^3$$ where $$s(1)=3$$

Integrate both sides w.r.t $t$, using the boundaries and replacing the dummy variables of integration:

$$\int_3^{s(t)}\,du=2\int_1^t \left(3v^2-1\right)^36v\,dv$$

Use a substitution on the RHS:

$$\int_3^{s(t)}\,du=2\int_2^{3t^2-1} w^3\,dw$$

Apply the FTOC:

$$s(t)-3=\frac{1}{2}\left(\left(3t^2-1\right)^4-2^4\right)$$

Or:

$$s(t)=\frac{1}{2}\left(3t^2-1\right)^4-5\quad\checkmark$$

 

[karush]

ok, guess I wasn't looking what forum i put it in