- Thread starter
- #1

#### karush

##### Well-known member

- Jan 31, 2012

- 3,055

$\tiny{2214.t1.13}$

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

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

Last edited: