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nadya68 [22]
3 years ago
13

First-order linear differential equations

Mathematics
1 answer:
kkurt [141]3 years ago
8 0

Answer:

(1)\ logy\ =\ -sint\ +\ c

(2)\ log(y+\dfrac{1}{2})\ =\ t^2\ +\ c

Step-by-step explanation:

1. Given differential equation is

  \dfrac{dy}{dt}+ycost = 0

=>\ \dfrac{dy}{dt}\ =\ -ycost

=>\ \dfrac{dy}{y}\ =\ -cost dt

On integrating both sides, we will have

  \int{\dfrac{dy}{y}}\ =\ \int{-cost\ dt}

=>\ logy\ =\ -sint\ +\ c

Hence, the solution of given differential equation can be given by

logy\ =\ -sint\ +\ c.

2. Given differential equation,

    \dfrac{dy}{dt}\ -\ 2ty\ =\ t

=>\ \dfrac{dy}{dt}\ =\ t\ +\ 2ty

=>\ \dfrac{dy}{dt}\ =\ 2t(y+\dfrac{1}{2})

=>\ \dfrac{dy}{(y+\dfrac{1}{2})}\ =\ 2t dt

On integrating both sides, we will have

   \int{\dfrac{dy}{(y+\dfrac{1}{2})}}\ =\ \int{2t dt}

=>\ log(y+\dfrac{1}{2})\ =\ 2.\dfrac{t^2}{2}\ + c

=>\ log(y+\dfrac{1}{2})\ =\ t^2\ +\ c

Hence, the solution of given differential equation is

log(y+\dfrac{1}{2})\ =\ t^2\ +\ c

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