What is the solution for y^{''}+b*y=0 ?

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

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y^{''}+b*y=0

y^{^{''}} + b \cdot y = 0

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

Solution steps

y^{''} (t) + b \cdot y = 0

Solve linear ODE:

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

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

\int_{1}^{2} \frac{6}{x ^{3} + 4 x} d x

x \to - 5 lim (\frac{2}{( x - 5 ) ^{3}})

y^{^{'}} + x y - 3 = 0

t an g e n t 50 x - 5 x

l a pl a ce t r an s f or m 3 (d^{2} x)

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