y''-2y'+1=0; y(0)=0, y'(0)=-1

Expert Solution

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Colby Messinger answered 11 months ago

y''' + 2y'' − y' − 2y = sin(4t), y(0) = 0, y'(0) = 0, y''(0) = 1

Given the differential equation y''+y'+2y=0, y(0)=−1, y'(0)=2y′′+y′+2y=0, y(0)=-1, y′(0)=2 Apply the Laplace Transform and solve for Y(s)=L{y}Y(s)=L{y}. You do not...

Given the differential equation y''+y'+2y=0, y(0)=−1, y'(0)=2y′′+y′+2y=0, y(0)=-1, y′(0)=2 Apply the Laplace Transform and solve for Y(s)=L{y}Y(s)=L{y}. You do not need to actually find the solution to the differential equation.

Use the Laplace transform to solve the problem with initial values y''-2y'+2y=cost y(0)=1 y'(0)=0

a) Solve IVP: y" + y' -2y = x + sin2x; y(0) = 1, y'(0) = 0

a) Solve IVP: y" + y' -2y = x + sin2x; y(0) = 1, y'(0) = 0 b) Solve using variation of parameters: y" -9y = x/e^3x

solve using the laplace transform y''-2y'+y=e^-1 , y(0)=0 , y'(0)=1

Solve the following problems: (a) y'' - 2y' + 5y = 0 with y(0) = 1...

Solve the following problems: (a) y'' - 2y' + 5y = 0 with y(0) = 1 and y'(0) = 2. (b) y(3) - 3y' + 2y = 0 with y(0) = 5, y'(0) = 6, and y''(0) = 11.

y''-3y+2y=e^3t y(0)=1 y'(0)=0 laplace transformation

solve using undetermined coefficients. y''' + 2y'' -y'-2y= 24e^-3t + 48t^2; y(0)=y'(0)=y''(0)=0

la place transform of 1. y´´+4y´+3y=0. y(0)=3, y´(0)=1 2. y´´+2y´+y=0. y(0)=1, y´(0)=1

Question : y''=4y'+8y=0 , (y''-4y'+13y)^2=0 , (y''+2y'+2y)^2=0 , y''-6y'+13y=0,y(0)=3 , y'(0)=13 , 2y''-6y'+17y=0,y(0)=2, y'(0)=13

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