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

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

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

y + y^{^{'}} = sin (t), y (π) = 1

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

Solution steps

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

Solve linear ODE:

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

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

\int_{1}^{3 x} (t^{2} + t) d t

f (x) = 2 - 2 x^{2}, - 16, at x = 3

x \to - 4 lim (10 x^{2} + 34 x - 24)

n = 1 \sum \infty 2 - 5 (0.9)^{n}

n = 0 \sum \infty \frac{1}{n} - \frac{1}{n + 1}

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