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Popular >

y^{'''}-2y^{''}-y^'+2y=e^4x

Solution

y^{^{'''}} - 2 y^{^{''}} - y^{^{'}} + 2 y = e^{4} x

Solution

y = c_{1} e^{2 x} + c_{2} e^{- x} + c_{3} e^{x} + \frac{e ^{4} x}{2} + \frac{e ^{4}}{4}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} e^{- x} + c_{3} e^{x} + \frac{e ^{4} x}{2} + \frac{e ^{4}}{4}

y = c_{1} e^{2 x} + c_{2} e^{- x} + c_{3} e^{x} + \frac{e ^{4} x}{2} + \frac{e ^{4}}{4}

Graph

Popular Examples

y^{''}+9y^'=sin(t),y(0)=1,y^'(0)=1

y^{^{''}} + 9 y^{^{'}} = sin (t), y (0) = 1, y^{^{'}} (0) = 1

y^{''}+9y=5sin(3x),y(0)=6,y^'(0)=0

y^{^{''}} + 9 y = 5 sin (3 x), y (0) = 6, y^{^{'}} (0) = 0

y^{''}+10y^'+25y=(-14e^{-5t})/(t^2+1)

y^{^{''}} + 10 y^{^{'}} + 25 y = \frac{- 14 e ^{- 5 t}}{t ^{2} + 1}

y^{''}+5y^'=sin(x)-cos(2x)

y^{^{''}} + 5 y^{^{'}} = sin (x) - cos (2 x)

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

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

Frequently Asked Questions (FAQ)

What is the solution for y^{'''}-2y^{''}-y^'+2y=e^4x ?
The solution for y^{'''}-2y^{''}-y^'+2y=e^4x is y=c_{1}e^{2x}+c_{2}e^{-x}+c_{3}e^x+(e^4x)/2+(e^4)/4