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Popular >

4y^{''}+4y^'+1=0

Solution

4 y^{^{''}} + 4 y^{^{'}} + 1 = 0

Solution

y = c_{1} + c_{2} e^{- t} - \frac{t}{4}

Solution steps

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

Solve linear ODE:

y = c_{1} + c_{2} e^{- t} - \frac{t}{4}

y = c_{1} + c_{2} e^{- t} - \frac{t}{4}

Popular Examples

D (x) = a - 2

(D^3+1)y=3+e^{-x}+5e^{2x}

(D^{3} + 1) y = 3 + e^{- x} + 5 e^{2 x}

x^{''}+49x=39cos(6t)

x^{^{''}} + 49 x = 39 cos (6 t)

4y^{''}-8y^'+4y=e^x(3x+2)+2cos(x)

4 y^{^{''}} - 8 y^{^{'}} + 4 y = e^{x} (3 x + 2) + 2 cos (x)

y^{''}+25y=3sin(5x)

y^{^{''}} + 25 y = 3 sin (5 x)

Frequently Asked Questions (FAQ)

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