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Popular >

y^{''}+3y^'+2y=x+6e^x,y(0)=0,y^'(0)=1

Solution

y^{^{''}} + 3 y^{^{'}} + 2 y = x + 6 e^{x}, y (0) = 0, y^{^{'}} (0) = 1

Solution

y = (- \frac{3}{4} - \frac{1}{4}) e^{- x} + \frac{3}{4} e^{- 2 x} + \frac{x}{2} - \frac{3}{4} + e^{x}

Solution steps

y^{''} (x) + 3 y^{'} (x) + 2 y = x + 6 e^{x}

Solve linear ODE:

y = (- \frac{3}{4} - \frac{1}{4}) e^{- x} + \frac{3}{4} e^{- 2 x} + \frac{x}{2} - \frac{3}{4} + e^{x}

y = (- \frac{3}{4} - \frac{1}{4}) e^{- x} + \frac{3}{4} e^{- 2 x} + \frac{x}{2} - \frac{3}{4} + e^{x}

Graph

Popular Examples

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y = sin (x y) + π, (1, π)

7(d^2y)/(dx^2)+4(dy)/(dx)+9y=0

7 \frac{d ^{2} y}{d x ^{2}} + 4 \frac{d y}{d x} + 9 y = 0

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f (x) = x^{4} + 3 x^{2} - x, (1, 3)

integral from 0 to a of x

\int_{0}^{a} x d x

integral from 0 to 1 of 6x^2+4x-3+4e^{4x}

\int_{0}^{1} 6 x^{2} + 4 x - 3 + 4 e^{4 x} d x

Frequently Asked Questions (FAQ)

What is the solution for y^{''}+3y^'+2y=x+6e^x,y(0)=0,y^'(0)=1 ?
The solution for y^{''}+3y^'+2y=x+6e^x,y(0)=0,y^'(0)=1 is y=(-3/4-1/4)e^{-x}+3/4 e^{-2x}+x/2-3/4+e^x