What is the solution for y^{''}-8y^'+16y=24x+2,y(0)=2,y^'(0)=-1 ?

The solution for y^{''}-8y^'+16y=24x+2,y(0)=2,y^'(0)=-1 is y= 9/8 e^{4x}-7e^{4x}x+(3x)/2+7/8

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y^{''}-8y^'+16y=24x+2,y(0)=2,y^'(0)=-1

y^{^{''}} - 8 y^{^{'}} + 16 y = 24 x + 2, y (0) = 2, y^{^{'}} (0) = - 1

y = \frac{9}{8} e^{4 x} - 7 e^{4 x} x + \frac{3 x}{2} + \frac{7}{8}

Solution steps

y^{''} (x) - 8 y^{'} (x) + 16 y = 24 x + 2

Solve linear ODE:

y = \frac{9}{8} e^{4 x} - 7 e^{4 x} x + \frac{3 x}{2} + \frac{7}{8}

y = \frac{9}{8} e^{4 x} - 7 e^{4 x} x + \frac{3 x}{2} + \frac{7}{8}

y^{^{'''}} + y^{^{''}} - 5 y^{^{'}} + 3 y = e^{- t} + sin (t)

y^{^{''}} - y^{^{'}} + 100 y = 10 sin (10 t)

\frac{d ^{2} y}{d x ^{2}} + 100 y = 1000 sin (10 x)

y^{^{''}} - 2 y^{^{'}} + y = 4 x^{2} - 3 + x^{- 1} e^{x}

y^{^{''}} - y^{^{'}} + 324 y = 18 sin (18 t)

What is the solution for y^{''}-8y^'+16y=24x+2,y(0)=2,y^'(0)=-1 ?
The solution for y^{''}-8y^'+16y=24x+2,y(0)=2,y^'(0)=-1 is y= 9/8 e^{4x}-7e^{4x}x+(3x)/2+7/8