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A solution of y' = -y is the function y(x)= . . .

User Mariarita
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1 Answer

3 votes

Answer:


y(x) = e^(-x) +C

Explanation:

As per the question,

Given first-order differential equation is

y' = -y

That is,


(dy)/(dx) + y = 0


(dy)/(y) = -dx

On Integrating both side,


\int (dy)/(y) = \int -dx

As we know that,


\int (1)/(x) dx = ln\ x +C

Therefore,


\int (dy)/(y) = \int -dx


ln\ y = e^(-x) + C

Hence the required function is


y(x) = e^(-x) +C

User Thibault Seisel
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