Question

2y' =3y+1 ; y(1) = 0 .

Answer 1

2y' =3y+1 ; y(1) = 0 .

$2{dy \over dx} = 3y +1$

$\int{2 \over (3y +1)}dy=\int dx$

$2\int{1 \over (3y +1)}dy=\int dx$

${2 \over 3} ln(3y +1)= x + c$

$ln(3y +1)={3 \over 2}x+c$

$3y +1= e^{{3 \over 2}x + c}$

$3y + 1= e^{{3 \over 2}x}. e^c$

$e^c= A$

$3y + 1=A e^{{3 \over 2}x}$

where A and c are constants

the initial conditionw given are y(0)=1

$3(0) +`1=Ae^{{3 \over 2}}$

$A={1 \over e^{{3 \over 2}}}= 0.22$