What is the solution for 2y^{''}+10y=0,y(0)=1,y^'(0)=-2 ?

The solution for 2y^{''}+10y=0,y(0)=1,y^'(0)=-2 is y=cos(sqrt(5)t)-2/(sqrt(5))sin(sqrt(5)t)

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2y^{''}+10y=0,y(0)=1,y^'(0)=-2

2 y^{^{''}} + 10 y = 0, y (0) = 1, y^{^{'}} (0) = - 2

y = cos (5 t) - \frac{2}{5} sin (5 t)

Solution steps

2 y^{''} (t) + 10 y = 0

Solve linear ODE:

y = cos (5 t) - \frac{2}{5} sin (5 t)

y = cos (5 t) - \frac{2}{5} sin (5 t)

\frac{d ^{2} x}{d t ^{2}} + 3 \frac{d x}{d t} + 8 x = 0

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

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

y^{^{''}} - y^{^{'}} - 6 y = 0, y (0) = 1, y^{^{'}} (0) = 2

\frac{d ^{2} x}{d t ^{2}} = - 49 x

What is the solution for 2y^{''}+10y=0,y(0)=1,y^'(0)=-2 ?
The solution for 2y^{''}+10y=0,y(0)=1,y^'(0)=-2 is y=cos(sqrt(5)t)-2/(sqrt(5))sin(sqrt(5)t)