What is the solution for y^{''}-2y^'+y=0,y(0)=3x^4 ?

The solution for y^{''}-2y^'+y=0,y(0)=3x^4 is y=c_{1}e^x+c_{2}xe^x

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y^{''}-2y^'+y=0,y(0)=3x^4

y^{^{''}} - 2 y^{^{'}} + y = 0, y (0) = 3 x^{4}

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

Solution steps

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

Solve linear ODE:

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

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

8 y^{^{''}} + 64 y^{^{'}} + 128 y = 0, y (0) = 5, y^{^{'}} (0) = 8

x^{^{''''}} - c x = 0

2 y^{^{''}} + 8 y = 0, y (0) = 2, y^{^{'}} (0) = 8

4 y^{^{''}} + y^{^{'}} + 2 y = 0

y^{^{''}} + 14 y^{^{'}} + 40 y = 0, y (0) = 9, y^{^{'}} (0) = - 48

What is the solution for y^{''}-2y^'+y=0,y(0)=3x^4 ?
The solution for y^{''}-2y^'+y=0,y(0)=3x^4 is y=c_{1}e^x+c_{2}xe^x