Using laplace transforms, solve x" + 4x' + 6x = 1- e^t with the following initial conditions: x(0) = x'(0) = 1.

The solution to the differential equation is

Step-by-step explanation:

Applying Laplace Transform will help us solve differential equations in Algebraic ways to find particular  solutions, thus after applying Laplace transform and evaluating at the initial conditions we need to solve and apply Inverse Laplace transform to find the final answer.

Applying Laplace Transform

We can start applying Laplace at the given ODE

So we will get

Applying initial conditions and solving for X(s).

If we apply the initial conditions we get

Simplifying

Moving all terms that do not have X(s) to the other side

Factoring X(s) and moving the rest to the other side.

Partial fraction decomposition method.

In order to apply Inverse Laplace Transform, we need to separate the fractions into the simplest form, so we can apply partial fraction decomposition to the first 2 fractions. For the first one we have

So if we multiply both sides by the entire denominator we get

At this point we can find the value of A fast if we plug s = 0, so we get

So the value of A is

We can replace that on the previous equation and multiply all terms by 6

We can simplify a bit

And by comparing coefficients we can tell the values of B and C

So the separated fraction will be

We can repeat the process for the second fraction.

Multiplying by the entire denominator give us

We can plug the value of s = 1 to find A fast.

So we get

We can replace that on the previous equation and multiply all terms by 11

Simplifying

And by comparing coefficients we can tell the values of B and C.

So the separated fraction will be

So far replacing both expanded fractions on the solution

We can combine the fractions with the same denominator

Simplifying give us

Completing the square

One last step before applying the Inverse Laplace transform is to factor the denominators using completing the square procedure for this case, so we will have

We are adding half of the middle term but squared, so the first 3 terms become the perfect  square, that is

So we get

Notice that the denominator has (s+2) inside a square we need to match that on the numerator so we can add and subtract 2

Lastly we can split the fraction one more

Applying Inverse Laplace Transform.

Since all terms are ready we can apply Inverse Laplace transform directly to each term and we will get

the value of v.w= -12.

step-by-step explanation:

we are given three vector r,v and w as:

r = < 5, -5, -2> ; v = < 2, -8, -8> ; w = < -2, 6, -5>

we know that r,v and w could also be represented as:

r=5i-5j-2k

v=2i-8j-8k

and w= -2i+6j-5k

where i,j and k are the unit vectors of x,y and z-axis respectively.

also we know that it has the following property:

i.i=1 ; j.j=1; k.k=1

and i.=j.i=i.k=k.i=j.k=k.j=0

i.e. when we multiply two vectors or we find it's dot product we need to only multiply the components of x coordinates with each other; y with each other and z with each other.

hence, v.w= (2i-8j- -2i+6j-5k)

                  = 2×(-2)+(-8)×(6)+(-8)×(-5)

                  = -4-48+40

                  = -12

hence, v.w= -12