What is the solution for ty^'+2y=sin(t),y(pi/2)=3 ?

The solution for ty^'+2y=sin(t),y(pi/2)=3 is y=-(cos(t))/t+(sin(t))/(t^2)+(3pi^2-4)/(4t^2)

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ty^'+2y=sin(t),y(pi/2)=3

t y^{^{'}} + 2 y = sin (t), y (\frac{π}{2}) = 3

y = - \frac{cos ( t )}{t} + \frac{sin ( t )}{t ^{2}} + \frac{3 π ^{2} - 4}{4 t ^{2}}

Solution steps

t y^{'} (t) + 2 y = sin (t)

Solve linear ODE:

y = - \frac{cos ( t )}{t} + \frac{sin ( t )}{t ^{2}} + \frac{3 π ^{2} - 4}{4 t ^{2}}

y = - \frac{cos ( t )}{t} + \frac{sin ( t )}{t ^{2}} + \frac{3 π ^{2} - 4}{4 t ^{2}}

y^{^{'}} = y (3 - y t)

\frac{d}{d x} (3 \cdot sin (x))

2 x^{2} y^{^{''}} + 5 x y^{^{'}} + y = 18 x + 4

y^{^{'}} + y^{2} sin (x) = 0

x \to \infty lim (sin (1))

What is the solution for ty^'+2y=sin(t),y(pi/2)=3 ?
The solution for ty^'+2y=sin(t),y(pi/2)=3 is y=-(cos(t))/t+(sin(t))/(t^2)+(3pi^2-4)/(4t^2)