(D-4)[x]+6 y=9 e^-3 r ; x(0)=-9x-(D-1)[y]=5 e^-3 t ; y(0)=4

Name: Problem 7, Chapter 7: Fundamentals of Differential Equations
Uploaded: 2020-07-01T16:23:51-08:00
Duration: 6 min 41 s
Channel: Tanishq Gupta
Description: (D-4)[x]+6 y=9 e^-3 r ; x(0)=-9x-(D-1)[y]=5 e^-3 t ; y(0)=4

(D-4)[x]+6 y=9 e^-3 r ; x(0)=-9x-(D-1)[y]=5 e^-3 t ; ...

Key Concepts

Systems of Differential Equations

Systems of differential equations involve more than one differential equation that are interconnected through the dependent variables. In the given problem, the functions x and y are related by their respective differential equations, making it necessary to consider the system as a whole. Solving such systems often involves techniques like elimination, substitution, or using matrix methods to find the unique solution satisfying all the equations simultaneously.

Non-Homogeneous Differential Equations

A non-homogeneous differential equation features a forcing term, which is a function not identically zero, on the right-hand side of the equation. In the provided equations, the exponential terms act as forcing functions. This non-zero right-hand side requires the solution to have a particular part, in addition to the homogeneous solution, to satisfy the entire differential equation.

Initial Value Problem

An initial value problem (IVP) specifies the values of the unknown functions at a particular point in the domain, usually at the initial time, to ensure a unique solution. In the given problem, the initial conditions provided for the functions help determine the unique trajectory of the solution, which is a standard practice when solving differential equations.

Differential Operator

This concept refers to the use of an operator, typically denoted as D, which represents differentiation with respect to an independent variable. In the context of differential equations, operators like (D - 4) or (D - 1) are applied to functions, compactly representing their derivatives minus a multiple of the function itself. This facilitates working with linear differential equations using algebraic techniques rather than dealing directly with derivatives.

Linear Differential Equations

Linear differential equations are equations in which the dependent variable and all its derivatives appear to the first power and are not multiplied together. The given equations are examples of first-order linear differential equations where the linear differential operator is applied to the unknown functions. These equations may include constant coefficients, making them amenable to methods such as superposition and transforming techniques.

Transcript

00:01 In the equation, we have to find the lapis transform of the given linear equation, which is d minus 4 x plus 6y equals to 9e raised to the per minus 3 t at x 0 equals to minus 9.

00:20 So this equation would be x dash minus 4x plus xy equals 9 a raised to the par minus 3 t.

00:29 Let it be first.

00:30 Second equation given is x -d.

00:33 Minus d minus 1 y equals to 5a raised to the power minus 3 t at y at 0 equals to 4 and this equation would be x minus y dash plus y equals to 5b ratio to the bar minus 3 t let it be second equation now taking the la plus transform of equation when we get s x s minus x at 0 minus 4 x s plus 6 y s equals to 9 1 upon s plus 3 now as we know that x at 0 is minus 9 so putting the value as x s plus 9 plus 9 plus 6 y s equals to 9 upon s plus 3 solving this we would get s minus 4 x s plus x plus x plus x plus 9 s minus 18 upon s plus 3 lettered with third equation now taking the laplace transform of second equation we get x s minus s y s plus y at 0 plus y s equals to 5 upon s plus 3 now as we know that y at 0 is 4 so putting the value we get x s minus s y s plus 4 plus y s equals to 5 upon s plus 3 so the value would be x s plus one minus s y s equals minus four s minus seven upon s plus three solving the equation third and fourth we get x s equals to minus three three s square plus eleven s plus eight upon s plus three s plus three s square minus 5 s plus 10 using partial fraction we will get minus 150 s minus 126 upon 17 s minus 5 s plus 10 minus 3 upon 17 s plus 3 upon 17 s plus 3 that is 1 15 x x .6 .15 into s minus 5 by 2 upon s minus 5 by 2 square plus 15 by 4 minus 501 upon 17 into 1 1 upon s minus 1 upon s minus 5 by 2 square plus 15 by 4 minus 5.

03:55 3 by 17 is plus 3.

03:59 Now taking the laplace transform, we get xt equals to minus 150e 5 t by 2 upon 17, pause root 15 t upon 2 minus 3, 3, 4, root 15, 4 ,000, 5, e 5, 5 t by 2, 2, 2 upon 85 sine root 15 t by 2 minus 3 a ratio 1 minus 3 t by 17 so this is the value for x g now we would be finding y t so similarly y s would be equal to 4 s cube minus 22 s square minus 28 s plus 46 upon s minus 1 s plus 3 s square minus 5 s plus 10 using the partial fraction we would get 46 upon 17 s minus 5 by 2 s minus 5 by 2 s minus 5 by 2 whole square plus 15 by 4 minus 219 .19.

05:34 17 and 2 .1 .s.

05:39 Minus 5 upon 2 square plus 15 by 4 plus 22 by 17 s plus 3.

05:50 Now taking inverse laplace transform, we get y...

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