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Popular >

y^{''}-y^'=1+e^x

Solution

y^{^{''}} - y^{^{'}} = 1 + e^{x}

Solution

y = c_{1} e^{x} + c_{2} - x + e^{x} x

Solution steps

y^{''} (x) - y^{'} (x) = 1 + e^{x}

Solve linear ODE:

y = c_{1} e^{x} + c_{2} - x + e^{x} x

y = c_{1} e^{x} + c_{2} - x + e^{x} x

Graph

Popular Examples

y^{'''}-y^'=e^{5t}

y^{^{'''}} - y^{^{'}} = e^{5 t}

y^{''}+4y^'+4y=(x+3)*e^{(-2x)}

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

y^{''}+9y=e^{2t}\H(t-1)

y^{^{''}} + 9 y = e^{2 t} H (t - 1)

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

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

y^{''}-3y^'+2y=x+1-10cos(3x)

y^{^{''}} - 3 y^{^{'}} + 2 y = x + 1 - 10 cos (3 x)

Frequently Asked Questions (FAQ)

What is the solution for y^{''}-y^'=1+e^x ?
The solution for y^{''}-y^'=1+e^x is y=c_{1}e^x+c_{2}-x+e^xx