Use the Laplace transform to solve the given initial value problem. y^w-y=0 ; y(0)=1, y^'(0)

step 1 So our problem here is y to the 4 prime minus y equals 0. We get at that y is 0 equals 1, y prim

Use the Laplace transform to solve the given initial value...

Key Concepts

Transformation of Derivatives

One of the key properties of the Laplace transform is its ability to convert derivatives into polynomial expressions involving the transform variable s. For example, taking the Laplace transform of a derivative converts it to an expression that includes the initial value of the function, streamlining the process of solving higher-order differential equations by reducing them to algebraic equations.

Inverse Laplace Transform

After solving the algebraic equation in the Laplace domain, the inverse Laplace transform is used to convert the solution back into the time domain. This step is essential in obtaining the explicit time-dependent solution for the original differential equation, completing the process of solving the initial value problem.

Laplace Transform

The Laplace transform is an integral transform that converts a function of time into a function of a complex variable, usually denoted as s. This tool simplifies the process of solving linear differential equations by turning differentiation and integration operations into algebraic manipulations, making it easier to handle initial conditions and transform complex differential equations into simpler forms.

Initial Value Problems

An initial value problem involves a differential equation along with specified values of the function and its derivatives at a particular point, typically at time zero. This information is critical for ensuring a unique solution when applying techniques like the Laplace transform, as it allows for the accurate handling of initial conditions during the transformation process.

Transcript

00:01 So our problem here is y to the 4 prime minus y equals 0.

00:08 We get at that y is 0 equals 1, y prime 0 equals 1, and y triple prime 0 equals 1, and y true prime 0 to 0.

00:15 So to get start with this, we're going to start off taking a gloss transforms of each term.

00:20 We're going to s to the 4th.

00:22 That's the gloss transform of y minus s cubed, y to the 0, minus s.

00:34 Squared y prime of 0 minus s y double prime of 0 minus y triple prime of 0 minus y triple prime of 0 minus a plos transform of y equal 0 it's not we're going to cancel out terms so we see that y prime of 0 why triple prime of 0 are both equal to 0 so you can cross off all of these terms this is going to be 0 and this is going to be 0 and we see that these two terms is both equal 1, so we know that this term is just going to be equal to s, and this term is just going to be equal to s cubed.

01:21 So now we're going to rewrite it down here, simpler, s to the fourth, ploss transform of y, minus s cubed, minus s minus laplace transform of y, all equal to zero.

01:40 So now it's much simpler.

01:43 We're going to separate it out.

01:44 So now on the left hand side, we're going to have plus transform of y outside of s to the fourth minus one.

01:57 On the right hand side, we have s to the third plus s.

02:03 So we're going to divide to get the plus transform alone.

02:10 S to the third plus s all over s to the fourth minus one.

02:22 All right, so now we have on top, we're going to try to simplify this out so we can do something with it.

02:34 And so we're going to step out of our denominator first.

02:37 So we see this is a one is the other one squared.

02:40 So we see this is a perfect square or it's a product of a s plus one and s minus one.

02:47 I figure what that's called.

02:48 But this is going to be s cubed plus three up top.