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Popular >

y^{''}+y^'+2=0

Solution

y^{^{''}} + y^{^{'}} + 2 = 0

Solution

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

Solution steps

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

Solve linear ODE:

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

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

Popular Examples

solvefor d,d(r)=(sqrt(r))/pi-2r^2 solve for

d, d (r) = \frac{r}{π} - 2 r^{2}

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

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

y^{''}-4y^'+4y=5t^2

y^{^{''}} - 4 y^{^{'}} + 4 y = 5 t^{2}

(d^2y)/(dx^2)-16(dy)/(dx)+64y=(x+1)e^x

\frac{d ^{2} y}{d x ^{2}} - 16 \frac{d y}{d x} + 64 y = (x + 1) e^{x}

y^{''}-5y^'+2y=xe^x

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

Frequently Asked Questions (FAQ)

What is the solution for y^{''}+y^'+2=0 ?
The solution for y^{''}+y^'+2=0 is y=c_{1}+c_{2}e^{-t}-2t