What is the solution for y^{''}+y=0,y(pi/4)=0,y^'(pi/4)=-3 ?

The solution for y^{''}+y=0,y(pi/4)=0,y^'(pi/4)=-3 is y= 3/(sqrt(2))cos(t)-3/(sqrt(2))sin(t)

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y^{''}+y=0,y(pi/4)=0,y^'(pi/4)=-3

y^{^{''}} + y = 0, y (\frac{π}{4}) = 0, y^{^{'}} (\frac{π}{4}) = - 3

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

Solution steps

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

Solve linear ODE:

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

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

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

\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

What is the solution for y^{''}+y=0,y(pi/4)=0,y^'(pi/4)=-3 ?
The solution for y^{''}+y=0,y(pi/4)=0,y^'(pi/4)=-3 is y= 3/(sqrt(2))cos(t)-3/(sqrt(2))sin(t)