What is the solution for y^{''}-2y+y=0,y(0)=xe^x ?

The solution for y^{''}-2y+y=0,y(0)=xe^x is y=c_{1}e^x+c_{2}e^{-x}

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y^{''}-2y+y=0,y(0)=xe^x

y^{^{''}} - 2 y + y = 0, y (0) = x e^{x}

y = c_{1} e^{x} + c_{2} e^{- x}

Solution steps

y^{''} (x) - 2 y + y = 0

Solve linear ODE:

y = c_{1} e^{x} + c_{2} e^{- x}

y = c_{1} e^{x} + c_{2} e^{- x}

\frac{d ^{2} y}{d x ^{2}} + 10 \frac{d y}{d x} + 61 y = 0

77 y^{^{''''}} + 219 y^{^{''}} + 100 y = 0

y^{^{''}} - 12 y^{^{'}} + 36 y = 0, y (0) = 5, y^{^{'}} (0) = 0

y^{^{''}} + 3 y = 4 y^{^{'}}

y^{^{'''}} + 5 y^{^{''}} - 34 y^{^{'}} - 80 y = 0

What is the solution for y^{''}-2y+y=0,y(0)=xe^x ?
The solution for y^{''}-2y+y=0,y(0)=xe^x is y=c_{1}e^x+c_{2}e^{-x}