Solve the given initial value problem with the Laplace transform. y^''+2 y^'+y=t e^-2 t, y(0)=

Solve the given initial value problem with the Laplace...

Key Concepts

Inverse Laplace Transform

The inverse Laplace transform is the process of converting an algebraic expression in the frequency domain back into a function in the time domain. It involves techniques like partial fraction decomposition and standard transform pairs to determine the original function that satisfies the differential equation.

Partial Fraction Decomposition

Partial fraction decomposition is a method used to break down complex rational expressions into simpler fractions. This technique is invaluable when performing inverse Laplace transforms, as it allows the decomposition of complicated expressions into forms whose inverse transforms are readily available in standard tables.

Shifting Theorem

The shifting theorem, especially the first shifting property, states that multiplying a function by an exponential term in the time domain corresponds to a shift in the complex frequency domain. This theorem is critical when handling functions multiplied by exponentials in differential equations, allowing for easier transformation and inversion.

Laplace Transform

The Laplace transform is an integral transform used to convert differential equations in the time domain into algebraic equations in the complex frequency domain. This transformation simplifies the process of solving linear ordinary differential equations, especially when initial conditions are specified, by turning differentiation into multiplication by the complex variable.

Initial Value Problem

An initial value problem is a differential equation accompanied by specified values of the function and its derivatives at a starting point. These initial conditions ensure the uniqueness of the solution and are essential in determining the particular solution, especially when using techniques like the Laplace transform.

Transform of Derivatives

The Laplace transform has specific properties for derivatives which allow differentiation in the time domain to be represented as algebraic terms in the frequency domain. This property replaces derivatives with expressions involving the Laplace variable and initial conditions, hence simplifying the solution process of differential equations.

Transcript

00:01 Okay, so our problem here is y2 prime minus 2y prime plus 2y equals e to the negative t.

00:06 And what we're given is that y is 0 equals 0, y prime 0 equals 1.

00:10 So we're going to start taking a lot of phosph transfer of each term plus s squared the cross transform of y minus s y 0 minus y prime of 0.

00:22 Oops, not an even sign.

00:25 Uh, minus 2 s.

00:29 I'm going to distribute this negative 2.

00:31 So we're going to have.

00:37 Plus 2 y is 0 plus 2 plus 2 plus 2 plus plus plus transform y equals the plus transform of e to the negative t all right so now we have that y 0 is 0 so it's going to go 0 and this is going to go 0 and i'll just in a different color y of 0 y prime 0 is 1 so you know this is going to go to 1 all right so let's flipize down we're going to s squared the plus transform of y minus 1 minus 2 s laplace transform of y plus 2 leplostransform of y equals and so we know that the the plos transform of an exponential function is going to be 1 over s minus the coefficient on the t so this is going to be at 1 s minus negative so it's going to be s plus 1 all right so we're going to further separate this out, so we're going to have a plus transform of y outside of s squared minus 2s plus 2 and move the one to the other side.

01:55 So we're going to have one over s plus one plus one.

01:59 And so if we've given the one the same denominator, we're going to end up with one over, or one plus s plus one over s plus one.

02:08 So that's just equal to s plus two over s plus one.

02:14 So now to get the laplace transform alone on the left side, we have laplace transfer of y is equal to s plus 2 all over s plus 1 times s squared minus 2 s plus 2.

02:33 All right.

02:34 And so from here we're going to try to do partial fraction decomposition, which is going to be a over s plus 1 plus b s .s.

02:44 Like that you got messy um plus b s plus c all over s squared minus 2 s plus 2 so we're going to multiply a by every term on the right -hand denominator and b -s plus c by s plus 1 and what that gives us is going to be a s squared minus 2 -as plus 2a and then the product of those, bs plus c and s plus 1, is going to be b .s squared plus c s plus b s plus c.

03:38 And all of that is equal to our numerator, which is s plus 2.

03:44 So now we're going to separate out into like terms.

03:46 So we'll have a .s squared plus b .s squared is equal to zero.

03:52 I'll have the negative 2a s plus b s plus c s equal to s and we have that 2a plus c is equal to 2.

04:07 So we're going to divide out, divide divide divide by s and so we're left with a is equal to negative b, negative 2a plus b plus c equals 1 and 2a plus c equals 2 just double check in the work real quick all right awesome okay so now we're going to substitute and solve for these terms so i'm going to looking at this right this left -hand term in the middle or this left -hand equation in the middle so i'm going to swap a with negative b so it's going to become just 2b because a is equal a negative so if i have a negative b times negative two, i have a 2b plus b plus c equals 1.

05:27 So 3b plus c is equal to 1.

05:43 I'm actually going to go the other route now that i thought about it...

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