y'' - 8y = 4, y(1) = 9, y'(0) = 5

Expert Solution

Colby Messinger answered 1 year ago

Find the solution of the given initial value problem: y^(4)+2y′′′+y′′+8y′−12y=12sint+40e^−t y(0)=0, y′(0)=38/5, y′′(0)=4/5, y′′′(0)=−54/5

Solve the equation y'=5-8y with initial condition y'(0)=0

1. Find the solution of y" + 8y' = 896sin(8t) + 256cos(8t) with y(0) = 9...

1. Find the solution of y" + 8y' = 896sin(8t) + 256cos(8t) with y(0) = 9 and y'(0) = 9 y = ? 2. Find y as a function of x if y"' - 9y" + 18y' = 20e^x , y(0) = 30 , y'(0) = 23 , y"(0) = 17 y(x) = ?

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Question : y'''+4y' =0 , y'''-2y''+4y'-8y=0 , y'''-3y''+3y'-y=0 , y^4 -4y'''+6y''-4y+y=0 , y^4+6y''+9y=0 , y^6+y'''=0

Solve: y'''−7y''+8y'+16y=0 y(0)=1, y'(0)=−3, y''(0)=10y(0)=1, y′(0)=-3, y′′(0)=10 y(t)= ?

3 Solve the following IVP a) y"+8y'-9y=0 y(1)=1, y'(1)=0 b)y"+4y'+5y=0 y(0)=1, y'(0)=0 c) y"-6y'+9y=0 y(0)=0, y'(0)=2...

3 Solve the following IVP a) y"+8y'-9y=0 y(1)=1, y'(1)=0 b)y"+4y'+5y=0 y(0)=1, y'(0)=0 c) y"-6y'+9y=0 y(0)=0, y'(0)=2 d)y"+4y'+4y=0 y(-1)=2, y'(-1)=1

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Solve the differential equation. y'' − 8y' + 20y = te^t, y(0) = 0, y'(0) =...

Solve the differential equation. y'' − 8y' + 20y = te^t, y(0) = 0, y'(0) = 0 (Answer using fractions)

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