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Popular >

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

Solution

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

Solution

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

Solution steps

y^{''} (x) - y^{'} (x) - 2 y = 1 - x

Solve linear ODE:

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

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

Popular Examples

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

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

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

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

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

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

y^{''}+4y=t^2+6e^t,y(0)=0,y^'(0)=9

y^{^{''}} + 4 y = t^{2} + 6 e^{t}, y (0) = 0, y^{^{'}} (0) = 9

x^{''}+x=9cos(4t)

x^{^{''}} + x = 9 cos (4 t)

Frequently Asked Questions (FAQ)

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