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

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In summary: I am a student and find it very useful for my hw and quizzesIn summary, we are given an initial value problem where we need to solve for the function s(t) using integration and substitution. The final solution is s(t) = 1/2(3t^2-1)^4-5.
  • #1
karush
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$\tiny{2214.t1.13}$
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$\textsf{solve the initial value problem}$
\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
 
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  • #2
I moved this here since IVP's are a topic from the study of differential equations.

We are given:

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

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

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

Use a substitution on the RHS:

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

Apply the FTOC:

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

Or:

\(\displaystyle s(t)=\frac{1}{2}\left(3t^2-1\right)^4-5\quad\checkmark\)
 
  • #3
ok, guess I wasn't looking what forum i put it in
 

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

What is an IVP?

An IVP, or initial value problem, is a type of mathematical problem that involves finding a solution to a differential equation with a known initial condition.

What is the meaning of the notation "s(1)=3" in this equation?

The notation "s(1)=3" indicates that the initial condition for the variable s is s=3 when t=1. This is used in the overall solution process for the IVP.

What does \frac{ds}{dt} mean in this equation?

The notation \frac{ds}{dt} represents the derivative of the variable s with respect to the variable t. In other words, it represents the rate of change of s with respect to t.

How do you solve this IVP?

To solve this IVP, you would first use integration techniques to solve for the general solution of the differential equation \frac{ds}{dt}=12t(3t^2-1)^3. Then, you would use the initial condition s(1)=3 to find the specific solution to the IVP.

What are some real-world applications of this type of IVP?

IVPs can be used to model a variety of real-world phenomena, such as population growth, chemical reactions, and electrical circuits. In this specific equation, it could potentially be used to model the motion of an object under the influence of a force.

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