What is the solution for y^{''}-5y^'+4y=x,y(0)=0,y^'(0)=0 ?

The solution for y^{''}-5y^'+4y=x,y(0)=0,y^'(0)=0 is y=(1/3-5/16)e^{4x}-1/3 e^x+x/4+5/16

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y^{''}-5y^'+4y=x,y(0)=0,y^'(0)=0

y^{^{''}} - 5 y^{^{'}} + 4 y = x, y (0) = 0, y^{^{'}} (0) = 0

y = (\frac{1}{3} - \frac{5}{16}) e^{4 x} - \frac{1}{3} e^{x} + \frac{x}{4} + \frac{5}{16}

Solution steps

y^{''} (x) - 5 y^{'} (x) + 4 y = x

Solve linear ODE:

y = (\frac{1}{3} - \frac{5}{16}) e^{4 x} - \frac{1}{3} e^{x} + \frac{x}{4} + \frac{5}{16}

y = (\frac{1}{3} - \frac{5}{16}) e^{4 x} - \frac{1}{3} e^{x} + \frac{x}{4} + \frac{5}{16}

\frac{d ^{2} y}{d t ^{2}} - 2 \frac{d y}{d t} - 35 y = (t + 1) e^{t}

\frac{d ^{2} y}{d x ^{2}} - 6 \frac{d y}{d x} + 8 y = 4 e^{3 x} + 8 x

y^{^{''''}} + 2 y^{^{''}} + y = cos (3 x) + 2

y^{^{''''}} + 2 y^{^{''}} + y = cos (3 x) + 3

y^{^{'''}} - 2 y^{^{''}} + y^{^{'}} = 3 cos (2 t)

What is the solution for y^{''}-5y^'+4y=x,y(0)=0,y^'(0)=0 ?
The solution for y^{''}-5y^'+4y=x,y(0)=0,y^'(0)=0 is y=(1/3-5/16)e^{4x}-1/3 e^x+x/4+5/16