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Popular >

y^{''}+4y^'+5y=1,y(0)=0,y^'(0)=0

Solution

y^{^{''}} + 4 y^{^{'}} + 5 y = 1, y (0) = 0, y^{^{'}} (0) = 0

Solution

y = e^{- 2 t} (- \frac{1}{5} cos (t) - \frac{2}{5} sin (t)) + \frac{1}{5}

Solution steps

y^{''} (t) + 4 y^{'} (t) + 5 y = 1

Solve linear ODE:

y = e^{- 2 t} (- \frac{1}{5} cos (t) - \frac{2}{5} sin (t)) + \frac{1}{5}

y = e^{- 2 t} (- \frac{1}{5} cos (t) - \frac{2}{5} sin (t)) + \frac{1}{5}

Graph

Popular Examples

10y^{''}+45y^'+2500y=180cos(7pix)

10 y^{^{''}} + 45 y^{^{'}} + 2500 y = 180 cos (7 π x)

y^{''}+4y=sin(t),y(0)=1,y^'(0)=-1

y^{^{''}} + 4 y = sin (t), y (0) = 1, y^{^{'}} (0) = - 1

y^{''}-4y=5e^{2x}+6,y^'(0)=-1,y(0)=0

y^{^{''}} - 4 y = 5 e^{2 x} + 6, y^{^{'}} (0) = - 1, y (0) = 0

y^{'''}+y^{''}-2y=xe^x+15

y^{^{'''}} + y^{^{''}} - 2 y = x e^{x} + 15

(d^2y)/(dx^2)-(dy)/(dx)+1/4 y=xe^{x/2}

\frac{d ^{2} y}{d x ^{2}} - \frac{d y}{d x} + \frac{1}{4} y = x e^{\frac{x}{2}}

Frequently Asked Questions (FAQ)

What is the solution for y^{''}+4y^'+5y=1,y(0)=0,y^'(0)=0 ?
The solution for y^{''}+4y^'+5y=1,y(0)=0,y^'(0)=0 is y=e^{-2t}(-1/5 cos(t)-2/5 sin(t))+1/5