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Popular >

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

Solution

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

Solution

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

Solution steps

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

Solve linear ODE:

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

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

Popular Examples

x^{''}-9x=2t^2-cos(t)

x^{^{''}} - 9 x = 2 t^{2} - cos (t)

y^{''}+4y^'+4y=e^{-t}

y^{^{''}} + 4 y^{^{'}} + 4 y = e^{- t}

3y^'+2y^{''}=-10cos(3.14*t/(0.2))

3 y^{^{'}} + 2 y^{^{''}} = - 10 cos (3.14 \cdot \frac{t}{0.2})

y^{''}+36y=sin(4x)

y^{^{''}} + 36 y = sin (4 x)

y^{''}+3y=-48x^2*e^{3x}

y^{^{''}} + 3 y = - 48 x^{2} \cdot e^{3 x}

Frequently Asked Questions (FAQ)

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