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y''-2y'+y=exx-1
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Math
y''-2y'+y=exx-1
y''-2y'+y=e
x
x
-1
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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
y''-2y'+1=0; y(0)=0, y'(0)=-1
y''-2y'+1=0; y(0)=0, y'(0)=-1
y''+2y'+2y=0,y(45)=2,y'(45)=-2
y''+2y'+2y=0,y(45)=2,y'(45)=-2
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
Use the Laplace transform to solve the problem with initial values y''-2y'+2y=cost y(0)=1 y'(0)=0
y"-y'-2y=54xe^2x
y"-y'-2y=54xe^2x
y"-2y'+2y = x^2+e^2x
y"-2y'+2y = x^2+e^2x
y''-3y'+2y=1+cost+e^-t
y''-3y'+2y=1+cost+e^-t
solve using the laplace transform y''-2y'+y=e^-1 , y(0)=0 , y'(0)=1
solve using the laplace transform y''-2y'+y=e^-1 , y(0)=0 , y'(0)=1
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
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