What is the solution for y^{''}+10y^'+25y=0,y(0)=0,y(pi/2)=1 ?

The solution for y^{''}+10y^'+25y=0,y(0)=0,y(pi/2)=1 is y=(2e^{(5pi-10t)/2}t)/(pi)

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y^{''}+10y^'+25y=0,y(0)=0,y(pi/2)=1

y^{^{''}} + 10 y^{^{'}} + 25 y = 0, y (0) = 0, y (\frac{π}{2}) = 1

y = \frac{2 e ^{\frac{5 π - 10 t}{2}} t}{π}

Solution steps

y^{''} (t) + 10 y^{'} (t) + 25 y = 0

Solve linear ODE:

y = \frac{2 e ^{\frac{5 π - 10 t}{2}} t}{π}

y = \frac{2 e ^{\frac{5 π - 10 t}{2}} t}{π}

\frac{8}{( s - 1 ) ^{3}}

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

\int x^{3} x^{2} - 16 d x

\frac{d}{d x} (5 - ∣ x ∣)

\int tan (3 x se) c^{2} 3 x d x

What is the solution for y^{''}+10y^'+25y=0,y(0)=0,y(pi/2)=1 ?
The solution for y^{''}+10y^'+25y=0,y(0)=0,y(pi/2)=1 is y=(2e^{(5pi-10t)/2}t)/(pi)