1 Answer

Poh · Answer 1 · 2020-01-25T10:30:56+0000

Answer:

Given differential equation,

(dy)/(dx)+5y=7

(dy)/(dx)=7-5y

$\implies (dy)/(7-5y)=dx$

Taking integration both sides,

$\int (dy)/(7-5y)=\int dx$

Put 7 - 5y = u ⇒ -5 dy = du ⇒ dy = -du/5,

$-(1)/(5) \int (du)/(u) = \log x + C$

$-(1)/(5) \log u = \log x + C$

$-(1)/(5)\log(7-5y) = \log x + C---(1)$

Here, x = 0, y = 0

$\implies -(1)/(5) \log 7= C$

Hence, from equation (1),

$-(1)/(5)\log(7-5y)=\log x -(1)/(5)log 7$

$\log(7-5y)=\log ((x)/(7^(1)/(5)))$

7-5y=(x)/(7^(1)/(5))

7-(x)/(7^(1)/(5))=5y

$\implies y=(1)/(5)(7-(x)/(7^(1)/(5)))$

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