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

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

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

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

y = c_{2} e^{t} sin (2 t)

Solution steps

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

Solve linear ODE:

y = c_{2} e^{t} sin (2 t)

y = c_{2} e^{t} sin (2 t)

\int 4 + x d x

\frac{d}{d x} (x^{2} + 2 x)

(y - x y^{2}) d x + x d y = 0

n = 1 \sum \infty \frac{1}{n ^{2} + n}

\frac{d}{d x} (3 x - 3)

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