What is the solution for 5y^{''}+20y=0,y(pi/4)=10,y(0)=15 ?

The solution for 5y^{''}+20y=0,y(pi/4)=10,y(0)=15 is y=15cos(2t)+10sin(2t)

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5y^{''}+20y=0,y(pi/4)=10,y(0)=15

5 y^{^{''}} + 20 y = 0, y (\frac{π}{4}) = 10, y (0) = 15

y = 15 cos (2 t) + 10 sin (2 t)

Solution steps

5 y^{''} (t) + 20 y = 0

Solve linear ODE:

y = 15 cos (2 t) + 10 sin (2 t)

y = 15 cos (2 t) + 10 sin (2 t)

2 y^{^{''}} + 6 y^{^{'}} = 0

y^{^{''}} + c^{2} y = 0, y (0) = a, y^{^{'}} (0) = 0

x^{^{''}} + 10 x^{^{'}} + 29 x = 0, x^{^{'}} (0) = 6

y^{^{''}} + 4 y = 0, y (0) = 1, y^{^{'}} (0) = 2

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

What is the solution for 5y^{''}+20y=0,y(pi/4)=10,y(0)=15 ?
The solution for 5y^{''}+20y=0,y(pi/4)=10,y(0)=15 is y=15cos(2t)+10sin(2t)