What is the solution for y^'+sin(x)y=sin(x),y(0)=pi ?

The solution for y^'+sin(x)y=sin(x),y(0)=pi is y=pie^{cos(x)-1}-e^{cos(x)-1}+1

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y^'+sin(x)y=sin(x),y(0)=pi

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

y = π e^{c o s (x) - 1} - e^{c o s (x) - 1} + 1

Solution steps

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

Solve separable ODE:

y = π e^{c o s (x) - 1} - e^{c o s (x) - 1} + 1

y = π e^{c o s (x) - 1} - e^{c o s (x) - 1} + 1

(87) (8 - 4)

\int \frac{15}{2} t + 8 e^{- t} d t

\int sec (u) d u

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

\frac{1}{2 r + 7 ^{\frac{6}{7}}}

What is the solution for y^'+sin(x)y=sin(x),y(0)=pi ?
The solution for y^'+sin(x)y=sin(x),y(0)=pi is y=pie^{cos(x)-1}-e^{cos(x)-1}+1