What is the solution for ty^'+2y=sin(t),y(pi/2)=1 ?

The solution for ty^'+2y=sin(t),y(pi/2)=1 is y=-(cos(t))/t+(sin(t))/(t^2)+(pi^2-4)/(4t^2)

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ty^'+2y=sin(t),y(pi/2)=1

t y^{^{'}} + 2 y = sin (t), y (\frac{π}{2}) = 1

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

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

f (x) = \frac{1}{x ^{\frac{2}{3}}}

\int \frac{4 x ^{3} + 3 x + 3}{x} d x

x \to 7 lim (sin (\frac{π}{x}))

\frac{\partial}{\partial y} (\frac{y}{x - 1})

y = e^{x^{3}}, (1, e)

What is the solution for ty^'+2y=sin(t),y(pi/2)=1 ?
The solution for ty^'+2y=sin(t),y(pi/2)=1 is y=-(cos(t))/t+(sin(t))/(t^2)+(pi^2-4)/(4t^2)