Answer:
(a) [tex]\frac{dy}{(2y+1)}=tdt[/tex] (b) [tex]y=\frac{e^{t^2}+e^{2c}-1}{2}[/tex]
Step-by-step explanation:
(1) We have given [tex]\frac{dy}{dt}+ycost=0[/tex]
[tex]\frac{dy}{dt}=-ycost[/tex]
[tex]\frac{dy}{y}=-costdt[/tex]
Integrating both side
[tex]lny=-sint+c[/tex]
[tex]y=e^{-sint}+e^{-c}[/tex]
(2) [tex]\frac{dy}{dt}-2ty=t[/tex]
[tex]\frac{dy}{dt}=2ty+t[/tex]
[tex]\frac{dy}{dt}=t(2y+1)[/tex]
[tex]\frac{dy}{(2y+1)}=tdt[/tex]
On integrating both side
[tex]\frac{ln(2y+1)}{2}=\frac{t^2}{2}+c[/tex]
[tex]ln(2y+1)={t^2}+2c[/tex]
[tex]2y+1=e^{t^2}+e^{2c}[/tex]
[tex]y=\frac{e^{t^2}+e^{2c}-1}{2}[/tex]
