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Popular >

x^{''}+x^'+4=0

Solution

x^{^{''}} + x^{^{'}} + 4 = 0

Solution

x = c_{1} + c_{2} e^{- t} - 4 t

Solution steps

x^{''} (t) + x^{'} (t) + 4 = 0

Solve linear ODE:

x = c_{1} + c_{2} e^{- t} - 4 t

x = c_{1} + c_{2} e^{- t} - 4 t

Graph

Popular Examples

y^{''}+4y=(sin(x))^3

y^{^{''}} + 4 y = (sin (x))^{3}

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

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

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

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

y^{'''}+4y^{''}-9y^'-36=x^2-4

y^{^{'''}} + 4 y^{^{''}} - 9 y^{^{'}} - 36 = x^{2} - 4

y^{''}+y= 1/((cos(x)))

y^{^{''}} + y = \frac{1}{( cos ( x ) )}

Frequently Asked Questions (FAQ)

What is the solution for x^{''}+x^'+4=0 ?
The solution for x^{''}+x^'+4=0 is x=c_{1}+c_{2}e^{-t}-4t