(i) Given
dxdy−y=3te2t with y(0)=1
Solving the differential equation,
Equation is linear in y and first order equation.
Integral factor will be e−∫dt=e−t
multiplying this factor on both sides of equation,
e−t(dxdy−y)=3te2te−t
e−t(dxdy−y)=3tet
Now integration both sides with respect to x,
e−ty=∫3tetdt
e−ty=3[tet−∫etdt]
e−ty=3[tet−et] +c
e−ty=3et(t−1)+c
so y=3e2t(t−1)+cet
applying condition, y(0)=1,
1=−3+c⟹c=4
so equation will be
y=3e2t(t−1)+4et
(ii) dxdy=(1−11x)y2 with y(0)=6−1
integrating both sides
∫y2dy=∫(1−11x)dx
y−1=x−211x2+c
applying condition y(0)=6−1
then 6=0−0+c⟹c=6
then equation will be
y−1=x−211x2+6
y−1=22x−11x2+12
then y=2x−11x2+12−2
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