What is the solution for y^{''}+4y^'-5=0 ?

The solution for y^{''}+4y^'-5=0 is y=c_{1}+c_{2}e^{-4t}+(5t)/4

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y^{''}+4y^'-5=0

y^{^{''}} + 4 y^{^{'}} - 5 = 0

y = c_{1} + c_{2} e^{- 4 t} + \frac{5 t}{4}

Solution steps

y^{''} (t) + 4 y^{'} (t) - 5 = 0

Solve linear ODE:

y = c_{1} + c_{2} e^{- 4 t} + \frac{5 t}{4}

y = c_{1} + c_{2} e^{- 4 t} + \frac{5 t}{4}

h \to 0 lim (\frac{h}{h ^{2} + 7 h + 1})

\frac{d}{d x} (\frac{1 - 4 x}{2 + x})

\frac{d y}{d x}, x y = y^{3}

\int \frac{2 v}{2 - 3 v ^{2}} d v

x y^{2} - 5 y + 6

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