What is the solution for 5y^{''}+(15)/(0.2)y=0,y(0)=0,y^'(0)=-10 ?

The solution for 5y^{''}+(15)/(0.2)y=0,y(0)=0,y^'(0)=-10 is y=-(2sqrt(5))/(sqrt(3))sin(sqrt(15)t)

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5y^{''}+(15)/(0.2)y=0,y(0)=0,y^'(0)=-10

5 y^{^{''}} + \frac{15}{0.2} y = 0, y (0) = 0, y^{^{'}} (0) = - 10

y = - \frac{2 5}{3} sin (15 t)

Solution steps

5 y^{''} (t) + \frac{15}{0.2} y = 0

Solve linear ODE:

y = - \frac{2 5}{3} sin (15 t)

y = - \frac{2 5}{3} sin (15 t)

\frac{d ^{2} y}{d x ^{2}} + 4 \frac{d y}{d x} + 7 y = 0

\frac{d ^{2} y}{d t ^{2}} + 2 \frac{d y}{d t} - 15 y = 0

(4 + 8 x) y^{^{''}} - 8 y + (4 - 8 x) y = 0

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

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

What is the solution for 5y^{''}+(15)/(0.2)y=0,y(0)=0,y^'(0)=-10 ?
The solution for 5y^{''}+(15)/(0.2)y=0,y(0)=0,y^'(0)=-10 is y=-(2sqrt(5))/(sqrt(3))sin(sqrt(15)t)