Question

First-order linear differential equations

1. dy/dt + ycost = 0 (Find the general solution)

2. dy/dt -2ty = t (Find the solution of the following IVP)

Answer 1

(a)
`(dy)/((2y+1))=tdt` (b)
`y=(e^(t^2)+e^(2c)-1)/(2)`

Explanation:

(1) We have given
`(dy)/(dt)+ycost=0`

`(dy)/(dt)=-ycost`

`(dy)/(y)=-costdt`

Integrating both side

`lny=-sint+c`

`y=e^(-sint)+e^(-c)`

(2)
`(dy)/(dt)-2ty=t`

`(dy)/(dt)=2ty+t`

`(dy)/(dt)=t(2y+1)`

`(dy)/((2y+1))=tdt`

On integrating both side

`(ln(2y+1))/(2)=(t^2)/(2)+c`

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

`2y+1=e^(t^2)+e^(2c)`

`y=(e^(t^2)+e^(2c)-1)/(2)`