What is the solution for y^{''}-2y^'+2y=0,y(pi)=6,y(0)=6 ?

The solution for y^{''}-2y^'+2y=0,y(pi)=6,y(0)=6 is y=e^t(-6/(e^pi)cos(t)+c_{2}sin(t))

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y^{''}-2y^'+2y=0,y(pi)=6,y(0)=6

y^{^{''}} - 2 y^{^{'}} + 2 y = 0, y (π) = 6, y (0) = 6

y = e^{t} (- \frac{6}{e ^{π}} cos (t) + c_{2} sin (t))

Solution steps

y^{''} (t) - 2 y^{'} (t) + 2 y = 0

Solve linear ODE:

y = e^{t} (- \frac{6}{e ^{π}} cos (t) + c_{2} sin (t))

y = e^{t} (- \frac{6}{e ^{π}} cos (t) + c_{2} sin (t))

y^{^{''}} + 9 y = 0, y (0) = 0, y (1) = 0

x^{^{''}} + 4.9 \cdot x = 0, x (0) = 0.087, x^{^{'}} (0) = 0

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

y^{^{''}} - 2 y^{^{'}} + 5 y = 0, y (0) = 1, y^{^{'}} (0) = - 3

\frac{1}{8} y^{^{''}} + 4 y^{^{'}} + 12 y = 0, y (0) = 0, y^{^{'}} (0) = 4

What is the solution for y^{''}-2y^'+2y=0,y(pi)=6,y(0)=6 ?
The solution for y^{''}-2y^'+2y=0,y(pi)=6,y(0)=6 is y=e^t(-6/(e^pi)cos(t)+c_{2}sin(t))