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

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

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

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

y = e^{2 t} (c_{1} cos (t) + c_{2} sin (t))

Solution steps

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

Solve linear ODE:

y = e^{2 t} (c_{1} cos (t) + c_{2} sin (t))

y = e^{2 t} (c_{1} cos (t) + c_{2} sin (t))

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

y^{^{'}} + y = 2 e^{x}

\frac{2}{x ^{\frac{1}{3}}} - 2

\int \frac{ln ( 4 x )}{2 x} d x

\int_{1}^{4} 7 \frac{ln ( x )}{x ^{2}} d x

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