**Differential Equation:**- $\dfrac{dr}{dt} = -ti – tj -tk $
**Initial Condition:**- $ r(0) = i + 2j +3k$

This problem aims to find the **initial value** of a vector function in the form of a differential equation. For this problem, one needs to understand the concept of initial values, **Laplace Transform**, and solve **differential equations** given the initial conditions.

An initial value problem, in **multivariable calculus**, is defined as a standard differential equation given with an** initial condition** that defines the value of the unknown function at a given point in a certain domain.

Now coming onto the **Laplace transform**, which is named after its creator Pierre Laplace, is an integral transform that transforms an arbitrary function of a real variable into a function of a **complex variable** $s$.

## Expert Answer:

Here, we have a simple** first-order derivative** and some initial conditions, so first we will be required to find a precise solution to this problem. One thing to note here is that the only condition we have will let us solve for the **one constant** we select when we integrate.

As we have defined above that if any problem is given to us as a derivative and with initial conditions to solve for an **explicit solution** is known as an initial value problem.

So we will start first by taking the **differential equation** and rearranging it for the value of $r$:

\[dr = (-ti – tj -tk) dt \]

**Integrating** on both sides:

\[ \int dr = \int(-ti – tj -tk) dt \]

Solving the Integral:

\[ r(t) = – \dfrac{t^2}{2}i – \dfrac{t^2}{2}j – \dfrac{t^2}{2}k + C \]

Putting the **initial condition** here $r(0)$:

\[ r(0) = 0i – 0j – 0k + C \]

One expression of $r(0)$ is given in question so we’re going to put both the **expressions** of $r(0)$ as equals:

\[ 0i – 0j – 0k + C = i + 2j +3k \]

$C$ comes out to be:

\[ C = i + 2j +3k \]

Now plugging $C$ back in $r$:

\[ r = – \dfrac{t^2}{2}i – \dfrac{t^2}{2}j – \dfrac{t^2}{2}k + C\]

\[ r = – \dfrac{t^2}{2}i – \dfrac{t^2}{2}j – \dfrac{t^2}{2}k + i + 2j +3k \]

## Numerical Result:

\[ r = – \left( \dfrac{t^2}{2} + 1\right)i – \left(\dfrac{t^2}{2}+2 \right)j – \left(\dfrac{t^2}{2}+3\right)k \]

## Example:

Solve the **initial value problem** for $r$ as a vector function of $t$.

**Differential Equation**:

\[\dfrac{dr}{dt} = -3ti – 3tj -tk \]

**Initial** Condition:

\[ r(0) = 2i + 4j +9k\]

**Rearranging** for $r$:

\[dr = (-3ti – 3tj -tk) dt \]

**Integrating** on both sides:

\[\int dr = \int(-3ti -3tj -tk) dt \]

Solving the Integral:

\[r = – \dfrac{-3t^2}{2}i – \dfrac{-3t^2}{2}j – \dfrac{t^2}{2}k + C \]

Putting $r(0)$:

\[ r(0) = 0i – 0j – 0k + C \]

Putting both **expressions** of $r(0) equals:$

\[ 0i – 0j – 0k + C = 2i + 4j +9k\]

$C$ comes out to be:

\[ C = 2i + 4j +9k \]

Now plugging $C$ back in $r$:

\[ r = – \dfrac{-3t^2}{2}i – \dfrac{-3t^2}{2}j – \dfrac{t^2}{2}k + 2i + 4j +9k \]

\[ r = \left( 2 – \dfrac{3t^2}{2}\right)i + \left( 4 -\dfrac{3t^2}{2} \right)j + \left(9 – \dfrac{t^2}{2}\right)k \]