What is the solution for y^{''}-4y^'+8y=0,y(0)=9,y(pi/4)=7 ?

The solution for y^{''}-4y^'+8y=0,y(0)=9,y(pi/4)=7 is y=e^{2t}(9cos(2t)+7/(e^{pi/2)}sin(2t))

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y^{''}-4y^'+8y=0,y(0)=9,y(pi/4)=7

y^{^{''}} - 4 y^{^{'}} + 8 y = 0, y (0) = 9, y (\frac{π}{4}) = 7

y = e^{2 t} (9 cos (2 t) + \frac{7}{e ^{\frac{π}{2}}} sin (2 t))

Solution steps

y^{''} (t) - 4 y^{'} (t) + 8 y = 0

Solve linear ODE:

y = e^{2 t} (9 cos (2 t) + \frac{7}{e ^{\frac{π}{2}}} sin (2 t))

y = e^{2 t} (9 cos (2 t) + \frac{7}{e ^{\frac{π}{2}}} sin (2 t))

What is the solution for y^{''}-4y^'+8y=0,y(0)=9,y(pi/4)=7 ?
The solution for y^{''}-4y^'+8y=0,y(0)=9,y(pi/4)=7 is y=e^{2t}(9cos(2t)+7/(e^{pi/2)}sin(2t))