What is the solution for y^{''}-6y^'+8y=x,y(0)=0,y^'(0)=0 ?

The solution for y^{''}-6y^'+8y=x,y(0)=0,y^'(0)=0 is y=(1/8-3/32)e^{4x}-1/8 e^{2x}+x/8+3/32

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y^{''}-6y^'+8y=x,y(0)=0,y^'(0)=0

y^{^{''}} - 6 y^{^{'}} + 8 y = x, y (0) = 0, y^{^{'}} (0) = 0

y = (\frac{1}{8} - \frac{3}{32}) e^{4 x} - \frac{1}{8} e^{2 x} + \frac{x}{8} + \frac{3}{32}

Solution steps

y^{''} (x) - 6 y^{'} (x) + 8 y = x

Solve linear ODE:

y = (\frac{1}{8} - \frac{3}{32}) e^{4 x} - \frac{1}{8} e^{2 x} + \frac{x}{8} + \frac{3}{32}

y = (\frac{1}{8} - \frac{3}{32}) e^{4 x} - \frac{1}{8} e^{2 x} + \frac{x}{8} + \frac{3}{32}

What is the solution for y^{''}-6y^'+8y=x,y(0)=0,y^'(0)=0 ?
The solution for y^{''}-6y^'+8y=x,y(0)=0,y^'(0)=0 is y=(1/8-3/32)e^{4x}-1/8 e^{2x}+x/8+3/32