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Popular >

y^{''}-4y=cosh(2x)

Solution

y^{^{''}} - 4 y = cosh (2 x)

Solution

y = c_{1} e^{2 x} + c_{2} e^{- 2 x} + \frac{1}{32} e^{2 x} (8 e^{- 2 x} x cosh (2 x) + 8 e^{- 2 x} x sinh (2 x) + 4 e^{- 2 x} sinh (2 x) + e^{- 4 x} - 4 x) + e^{- 2 x} \cdot \int - \frac{e ^{2 x} cosh ( 2 x )}{4} d x

Solution steps

y^{''} (x) - 4 y = cosh (2 x)

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} e^{- 2 x} + \frac{1}{32} e^{2 x} (8 e^{- 2 x} x cosh (2 x) + 8 e^{- 2 x} x sinh (2 x) + 4 e^{- 2 x} sinh (2 x) + e^{- 4 x} - 4 x) + e^{- 2 x} \cdot \int - \frac{e ^{2 x} cosh ( 2 x )}{4} d x

y = c_{1} e^{2 x} + c_{2} e^{- 2 x} + \frac{1}{32} e^{2 x} (8 e^{- 2 x} x cosh (2 x) + 8 e^{- 2 x} x sinh (2 x) + 4 e^{- 2 x} sinh (2 x) + e^{- 4 x} - 4 x) + e^{- 2 x} \cdot \int - \frac{e ^{2 x} cosh ( 2 x )}{4} d x

Popular Examples

y^{''}-3y^'=8e^{3x}+4sin(x)

y^{^{''}} - 3 y^{^{'}} = 8 e^{3 x} + 4 sin (x)

y^{''}+5y^'+6y=18t^2

y^{^{''}} + 5 y^{^{'}} + 6 y = 18 t^{2}

y^{''}+3y^'+2y=e^t,y(0)=1,y^'(0)=-6

y^{^{''}} + 3 y^{^{'}} + 2 y = e^{t}, y (0) = 1, y^{^{'}} (0) = - 6

y^{''}+2y^'+4y=111e^{2t}cos(3t)

y^{^{''}} + 2 y^{^{'}} + 4 y = 111 e^{2 t} cos (3 t)

y^{''}+4y= 1/(cos(2x))

y^{^{''}} + 4 y = \frac{1}{cos ( 2 x )}

Frequently Asked Questions (FAQ)

What is the solution for y^{''}-4y=cosh(2x) ?
The solution for y^{''}-4y=cosh(2x) is