What is the solution for y^{''}-2y^'+5y=0,y(pi)=0,y^'(pi)=2 ?

The solution for y^{''}-2y^'+5y=0,y(pi)=0,y^'(pi)=2 is y=e^{t-pi}sin(2t)

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

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

y = e^{t - π} sin (2 t)

Solution steps

y^{''} (t) - 2 y^{'} (t) + 5 y = 0

Solve linear ODE:

y = e^{t - π} sin (2 t)

y = e^{t - π} sin (2 t)

f (x) = 3 x^{2}, at x = 4

\frac{d}{d x} (\frac{x + 5}{x})

x^{2} y^{^{''}} + x y^{^{'}} - y = x^{\frac{1}{2}}

x \to \infty lim (\frac{4 ^{x} + 3}{4 ^{x + 1}})

f (x) = - \frac{2}{x}

What is the solution for y^{''}-2y^'+5y=0,y(pi)=0,y^'(pi)=2 ?
The solution for y^{''}-2y^'+5y=0,y(pi)=0,y^'(pi)=2 is y=e^{t-pi}sin(2t)