What is the solution for y^{''}-4y^'+4y=2e^{5x},y(0)=1,y^'(0)=1 ?

The solution for y^{''}-4y^'+4y=2e^{5x},y(0)=1,y^'(0)=1 is y= 7/9 e^{2x}-(5e^{2x}x)/3+2/9 e^{5x}

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y^{''}-4y^'+4y=2e^{5x},y(0)=1,y^'(0)=1

y^{^{''}} - 4 y^{^{'}} + 4 y = 2 e^{5 x}, y (0) = 1, y^{^{'}} (0) = 1

y = \frac{7}{9} e^{2 x} - \frac{5 e ^{2 x} x}{3} + \frac{2}{9} e^{5 x}

Solution steps

y^{''} (x) - 4 y^{'} (x) + 4 y = 2 e^{5 x}

Solve linear ODE:

y = \frac{7}{9} e^{2 x} - \frac{5 e ^{2 x} x}{3} + \frac{2}{9} e^{5 x}

y = \frac{7}{9} e^{2 x} - \frac{5 e ^{2 x} x}{3} + \frac{2}{9} e^{5 x}

y^{^{''}} - 2 y^{^{'}} + 5 y = 20 sin (t)

y^{^{''}} - 2 y^{^{'}} - 3 y = 6 x + 4

y^{^{''}} + 4 y = 6 sin (2 t), y (0) = 2, y^{^{'}} (0) = 1

10 y^{^{''}} + 1000 y = 160 cos (8 t)

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

What is the solution for y^{''}-4y^'+4y=2e^{5x},y(0)=1,y^'(0)=1 ?
The solution for y^{''}-4y^'+4y=2e^{5x},y(0)=1,y^'(0)=1 is y= 7/9 e^{2x}-(5e^{2x}x)/3+2/9 e^{5x}