Question

First-order linear differential equations

Answer 1

Answer:

(a) $\frac{dy}{(2y+1)}=tdt$ (b) $y=\frac{e^{t^2}+e^{2c}-1}{2}$

Step-by-step explanation:

(1) We have given $\frac{dy}{dt}+ycost=0$

$\frac{dy}{dt}=-ycost$

$\frac{dy}{y}=-costdt$

Integrating both side

$lny=-sint+c$

$y=e^{-sint}+e^{-c}$

(2) $\frac{dy}{dt}-2ty=t$

$\frac{dy}{dt}=2ty+t$

$\frac{dy}{dt}=t(2y+1)$

$\frac{dy}{(2y+1)}=tdt$

On integrating both side  

$\frac{ln(2y+1)}{2}=\frac{t^2}{2}+c$

$ln(2y+1)={t^2}+2c$

$2y+1=e^{t^2}+e^{2c}$

$y=\frac{e^{t^2}+e^{2c}-1}{2}$