What is the solution for y^{''}-2y+4y=0 ?

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

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y^{''}-2y+4y=0

y^{^{''}} - 2 y + 4 y = 0

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

Solution steps

y^{''} (t) - 2 y + 4 y = 0

Solve linear ODE:

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

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

\int 11 x^{10} - \frac{10}{x ^{10}} d x

\int \frac{1}{9 - 4 x ^{2}} d x

cos (2 x) \cdot ln (3 x - 5)

\frac{x ^{2}}{4 + x}

\frac{\partial}{\partial x} (6 e^{x^{5} y^{4}})

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