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Popular >

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

Solution

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

Solution

y = c_{1} cos (x) + c_{2} sin (x) + e^{x}

Solution steps

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

Solve linear ODE:

y = c_{1} cos (x) + c_{2} sin (x) + e^{x}

y = c_{1} cos (x) + c_{2} sin (x) + e^{x}

Graph

Popular Examples

y^{''}-y^'-6y=15e^{3t},y(0)=2,y^'(0)=-6

y^{^{''}} - y^{^{'}} - 6 y = 15 e^{3 t}, y (0) = 2, y^{^{'}} (0) = - 6

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

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

y^{''}+4y^'-5y=2e^{3x}

y^{^{''}} + 4 y^{^{'}} - 5 y = 2 e^{3 x}

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

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

y^{''}+2y^'+y= 1/6 e^{-x}

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

Frequently Asked Questions (FAQ)

What is the solution for y^{''}+y=2e^x ?
The solution for y^{''}+y=2e^x is y=c_{1}cos(x)+c_{2}sin(x)+e^x