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y''+2y'+2y=0,y(45)=2,y'(45)=-2
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y''+2y'+2y=0,y(45)=2,y'(45)=-2
y''+2y'+2y=0,y(45)=2,y'(45)=-2
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y''' + 2y'' − y' − 2y = sin(4t), y(0) = 0, y'(0) = 0, y''(0) = 1
y''' + 2y'' − y' − 2y = sin(4t), y(0) = 0, y'(0) = 0, y''(0) = 1
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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.
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Solve the initial value problem: y''+2y'+y = x^2 , y(0)=0 , y'(0) = 0
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y"-3y'+2y=4t-8 , y(0)=2 , y'(0)=7 y(t)=?
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