What is the solution for y^{''}-5y^'+4y=8e^x,y(0)=1,y^'(0)=1 ?

The solution for y^{''}-5y^'+4y=8e^x,y(0)=1,y^'(0)=1 is y=(1-1/9)e^{4x}+1/9 e^x-(8e^xx)/3

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y^{''}-5y^'+4y=8e^x,y(0)=1,y^'(0)=1

y^{^{''}} - 5 y^{^{'}} + 4 y = 8 e^{x}, y (0) = 1, y^{^{'}} (0) = 1

y = (1 - \frac{1}{9}) e^{4 x} + \frac{1}{9} e^{x} - \frac{8 e ^{x} x}{3}

Solution steps

y^{''} (x) - 5 y^{'} (x) + 4 y = 8 e^{x}

Solve linear ODE:

y = (1 - \frac{1}{9}) e^{4 x} + \frac{1}{9} e^{x} - \frac{8 e ^{x} x}{3}

y = (1 - \frac{1}{9}) e^{4 x} + \frac{1}{9} e^{x} - \frac{8 e ^{x} x}{3}

y^{^{''}} - 2 y^{^{'}} + 2 y = x + 1, y (0) = 3, y^{^{'}} (0) = 0

d, d (t) = 4000 - 3000 e - 0.06 t

y^{^{''}} - 2 y^{^{'}} + 2 y = e^{x} \frac{1}{sin ( x )}

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

\frac{d ^{2} x}{d t ^{2}} + 36 x = 10 sin (4 t), x (0) = 0

What is the solution for y^{''}-5y^'+4y=8e^x,y(0)=1,y^'(0)=1 ?
The solution for y^{''}-5y^'+4y=8e^x,y(0)=1,y^'(0)=1 is y=(1-1/9)e^{4x}+1/9 e^x-(8e^xx)/3