What is the solution for ty^'+2y=sin(t),y(pi/2)=5 ?

The solution for ty^'+2y=sin(t),y(pi/2)=5 is y=-(cos(t))/t+(sin(t))/(t^2)+(5pi^2-4)/(4t^2)

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ty^'+2y=sin(t),y(pi/2)=5

t y^{^{'}} + 2 y = sin (t), y (\frac{π}{2}) = 5

y = - \frac{cos ( t )}{t} + \frac{sin ( t )}{t ^{2}} + \frac{5 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{5 π ^{2} - 4}{4 t ^{2}}

y = - \frac{cos ( t )}{t} + \frac{sin ( t )}{t ^{2}} + \frac{5 π ^{2} - 4}{4 t ^{2}}

\frac{\partial}{\partial x} (6 x - 4 y)

\int (8 + x^{2}) d x

\int \frac{1}{8 - 2 x ^{2}} d x

\int \frac{2 x}{x ^{2} + 6 x + 5} d x

3 t cos (3 t)

What is the solution for ty^'+2y=sin(t),y(pi/2)=5 ?
The solution for ty^'+2y=sin(t),y(pi/2)=5 is y=-(cos(t))/t+(sin(t))/(t^2)+(5pi^2-4)/(4t^2)