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y^{''}-y^'-6y=-e^x+12x,y^'(0)=-2,y(0)=1

Solution

y^{^{''}} - y^{^{'}} - 6 y = - e^{x} + 12 x, y^{^{'}} (0) = - 2, y (0) = 1

Solution

y = \frac{e ^{3 x}}{6} + \frac{1}{3} e^{- 2 x} + \frac{1}{6} e^{x} - 2 x + \frac{1}{3}

Solution steps

y^{''} (x) - y^{'} (x) - 6 y = - e^{x} + 12 x

Solve linear ODE:

y = \frac{e ^{3 x}}{6} + \frac{1}{3} e^{- 2 x} + \frac{1}{6} e^{x} - 2 x + \frac{1}{3}

y = \frac{e ^{3 x}}{6} + \frac{1}{3} e^{- 2 x} + \frac{1}{6} e^{x} - 2 x + \frac{1}{3}

Graph

Popular Examples

2x^{''}+5x^'+2x=4e^{-2t}-t

2 x^{^{''}} + 5 x^{^{'}} + 2 x = 4 e^{- 2 t} - t

y^{''}-9y=(9t)/(e^{3t)}

y^{^{''}} - 9 y = \frac{9 t}{e ^{3 t}}

(d^2x)/(dt^2)-3(dx)/(dt)+12x=9t

\frac{d ^{2} x}{d t ^{2}} - 3 \frac{d x}{d t} + 12 x = 9 t

y^{''}+y^'+5y=2x

y^{^{''}} + y^{^{'}} + 5 y = 2 x

y^{''}-4y^'+13y=e^{2t}*sin(3t),y(0)=4

y^{^{''}} - 4 y^{^{'}} + 13 y = e^{2 t} \cdot sin (3 t), y (0) = 4

Frequently Asked Questions (FAQ)

What is the solution for y^{''}-y^'-6y=-e^x+12x,y^'(0)=-2,y(0)=1 ?
The solution for y^{''}-y^'-6y=-e^x+12x,y^'(0)=-2,y(0)=1 is y=(e^{3x})/6+1/3 e^{-2x}+1/6 e^x-2x+1/3