Solve the IVP by applying the Laplace transform:y''+y=sqrt(2)*sin[sqrt(2)t]; y(0)=10, y'(0)=0

Expert Solution

milcah answered 1 year ago

using the Laplace transform solve the IVP y'' +4y= 3sin(t) y(0) =1 , y'(0) = -...

using the Laplace transform solve the IVP y'' +4y= 3sin(t) y(0) =1 , y'(0) = - 1 , i am stuck on the partial fraction decomposition step. please explain the decomposition clearly.

Use the Laplace transform to solve the IVP: y^'''+y^''+3y^'-5y =16e^(-t); y(0)=0; y'(0)=2; y^'' (0)= -4

Use laplace transform to solve IVP 2y”+3y’+y=8e^(-2t) , y(0)=-4 , y’(0)=2

y'' - y = e^(-t) - (2)(t)(e^(-t)) y(0)= 1 y'(0)= 2 Use Laplace Transforms to solve....

y'' - y = e^(-t) - (2)(t)(e^(-t)) y(0)= 1 y'(0)= 2 Use Laplace Transforms to solve. Sketch the solution or use matlab to show the graph.

Solve the IVP using Laplace transforms x' + y'=e^t -x''+3x' +y =0 x(0)=0, x'(0)=1, y(0)=0

Take the Laplace transform of the following initial value and solve for Y(s)=L{y(t)}: y′′+4y={sin(πt) ,0, 0≤t<11≤t...

Take the Laplace transform of the following initial value and solve for Y(s)=L{y(t)}: y′′+4y={sin(πt) ,0, 0≤t<11≤t y(0)=0,y′(0)=0 Y(s)= ? Hint: write the right hand side in terms of the Heaviside function. Now find the inverse transform to find y(t). Use step(t-c) for the Heaviside function u(t−c) . y(t)= ?

solve y'' + 4y =6 sin (t);y(0) =6, y'(0)=0 using 1st laplace transforms, 2nd undertinend coefficent,...

solve y'' + 4y =6 sin (t);y(0) =6, y'(0)=0 using 1st laplace transforms, 2nd undertinend coefficent, and 3rd variation of parameters.

Solve y''-y=e^t(2cos(t)-sin(t)) with initial condition y(0)=y'(0)=0

Please draw the solution without solve the IVP y"+y=dirac function (t-pi/2) y(0)=0, y'(0) . (Label y(t)...

Please draw the solution without solve the IVP y"+y=dirac function (t-pi/2) y(0)=0, y'(0) . (Label y(t) and t number as well) I need a professional expert to answer this question. (be able to follow the comment) (Show the step for what you need to get for drawing this solution as well)

SOLVE the IVP: (D^2+1)y = e^t, y(0) = -1 and y'(0) = 1. Thank you.

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