What is the solution for 2y^{''}+5y^'-3y=0 ?

The solution for 2y^{''}+5y^'-3y=0 is y=c_{1}e^{t/2}+c_{2}e^{-3t}

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2y^{''}+5y^'-3y=0

2 y^{^{''}} + 5 y^{^{'}} - 3 y = 0

y = c_{1} e^{\frac{t}{2}} + c_{2} e^{- 3 t}

Solution steps

2 y^{''} (t) + 5 y^{'} (t) - 3 y = 0

Solve linear ODE:

y = c_{1} e^{\frac{t}{2}} + c_{2} e^{- 3 t}

y = c_{1} e^{\frac{t}{2}} + c_{2} e^{- 3 t}

\frac{\partial}{\partial y} (2 x y + z^{2})

\frac{\partial}{\partial x} (3 x ln (y) + y ln (z) - 2 x)

\int x (x + \frac{1}{x}) d x

\int x^{2} d x

x \to 0 + lim (\frac{2 x}{∣ x ∣})

What is the solution for 2y^{''}+5y^'-3y=0 ?
The solution for 2y^{''}+5y^'-3y=0 is y=c_{1}e^{t/2}+c_{2}e^{-3t}