What is the solution for y^{''}+25y=0,y(0)=1,y^'(0)=-1 ?

The solution for y^{''}+25y=0,y(0)=1,y^'(0)=-1 is y=cos(5t)-1/5 sin(5t)

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y^{''}+25y=0,y(0)=1,y^'(0)=-1

y^{^{''}} + 25 y = 0, y (0) = 1, y^{^{'}} (0) = - 1

y = cos (5 t) - \frac{1}{5} sin (5 t)

Solution steps

y^{''} (t) + 25 y = 0

Solve linear ODE:

y = cos (5 t) - \frac{1}{5} sin (5 t)

y = cos (5 t) - \frac{1}{5} sin (5 t)

y^{^{''}} - 18 y^{^{'}} + 97 y = 0

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

y^{^{''}} + \frac{16}{3} y = 0

4 y^{^{''}} + 22 y^{^{'}} + 18 y = 0, y (0) = 3, y^{^{'}} (0) = 0

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

What is the solution for y^{''}+25y=0,y(0)=1,y^'(0)=-1 ?
The solution for y^{''}+25y=0,y(0)=1,y^'(0)=-1 is y=cos(5t)-1/5 sin(5t)