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Popular >

y^{''}-y^'-2y=2cos(2x)+sin(2x)

Solution

y^{^{''}} - y^{^{'}} - 2 y = 2 cos (2 x) + sin (2 x)

Solution

y = c_{1} e^{2 x} + c_{2} e^{- x} - \frac{1}{4} sin (2 x) - \frac{1}{4} cos (2 x)

Solution steps

y^{''} (x) - y^{'} (x) - 2 y = 2 cos (2 x) + sin (2 x)

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} e^{- x} - \frac{1}{4} sin (2 x) - \frac{1}{4} cos (2 x)

y = c_{1} e^{2 x} + c_{2} e^{- x} - \frac{1}{4} sin (2 x) - \frac{1}{4} cos (2 x)

Graph

Popular Examples

y^{''}+4y=-5,y(pi/8)=-1/4 ,y^'(pi/8)=2

y^{^{''}} + 4 y = - 5, y (\frac{π}{8}) = - \frac{1}{4}, y^{^{'}} (\frac{π}{8}) = 2

y^{''}+7y^'+6y=3sin(2t)

y^{^{''}} + 7 y^{^{'}} + 6 y = 3 sin (2 t)

y^{''}-2y^'+5y=2x^2+12

y^{^{''}} - 2 y^{^{'}} + 5 y = 2 x^{2} + 12

y^{''}+2y^'+y=5e^{-t}

y^{^{''}} + 2 y^{^{'}} + y = 5 e^{- t}

y^{''}+y=-cos(2t),y(0)=0,y^'(0)=0

y^{^{''}} + y = - cos (2 t), y (0) = 0, y^{^{'}} (0) = 0

Frequently Asked Questions (FAQ)

What is the solution for y^{''}-y^'-2y=2cos(2x)+sin(2x) ?
The solution for y^{''}-y^'-2y=2cos(2x)+sin(2x) is y=c_{1}e^{2x}+c_{2}e^{-x}-1/4 sin(2x)-1/4 cos(2x)