What is the solution for ty^'+2y=sin(t),y(pi/2)=9 ?

The solution for ty^'+2y=sin(t),y(pi/2)=9 is y=-(cos(t))/t+(sin(t))/(t^2)+(9pi^2-4)/(4t^2)

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ty^'+2y=sin(t),y(pi/2)=9

t y^{^{'}} + 2 y = sin (t), y (\frac{π}{2}) = 9

y = - \frac{cos ( t )}{t} + \frac{sin ( t )}{t ^{2}} + \frac{9 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{9 π ^{2} - 4}{4 t ^{2}}

y = - \frac{cos ( t )}{t} + \frac{sin ( t )}{t ^{2}} + \frac{9 π ^{2} - 4}{4 t ^{2}}

\int x^{2} ln (x) d x

\int_{2}^{7} t^{4} ln (3 t) d t

x^{- 9}

\int e^{r^{2}} r d r

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

What is the solution for ty^'+2y=sin(t),y(pi/2)=9 ?
The solution for ty^'+2y=sin(t),y(pi/2)=9 is y=-(cos(t))/t+(sin(t))/(t^2)+(9pi^2-4)/(4t^2)