Question

I have a question. I need step-by step how so solve this: y'+y=cos(t) y'(0)=0=y(0)

Answer 1

\[y \prime+1.y=\cos t\]
\[I.F.=e ^{\int\limits 1 dt}=e^t\]
C.S. is \[y.e^t=\int\limits \cos t ~e^t dt+c=I1+c\]
\[I1=\int\limits \cos t~ e^t dt=\cos t *e^t-\int\limits \left(- \sin t \right)e^t ~dt\]
\[=e^t \cos t+\left[ \sin t*e ^{t}-\int\limits \cos t*e^t ~dt \right]=e^t \left( \cos t+\sin t \right)-I1\]
\[2 I1=e^t \left( \cos t+\sin t \right)\]
\[I1=\frac{ 1 }{ 2 }e^t \left( \cos t+\sin t \right)\]
\[y*e^t=\frac{ 1 }{ 2 }e^t \left( \cos t+\sin t \right)+c\]
when t=0,y=0
\[0=\frac{ 1 }{ 2 }\left( \cos 0+\sin 0 \right)+c,0=\frac{ 1 }{ 2 }\left( 1+0 \right)+c,c=-\frac{ 1 }{ 2 }\]
substitute the value of c and simplify