The solution of the differential equation y'={1}{e^{-y}-x}, is

Correct Answer is : x=e−y(y+c)x=e^{-y}(y+c)x=e−y(y+c)

Correct Answer is : x=e−y(y+c)x=e^{-y}(y+c)x=e−y(y+c)

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TS EAMCET 2016 Code A Solved Paper

Section: Mathematics

Question : 80 of 160

Marks: +1, -0

y'=\frac{1}{e^{-y}-x}

, is

Solution:

The given differential equation is,

y'=\;\frac{1}{e^{-y}-x}

\;\frac{dy}{dx}=\;\frac{1}{e^{-y}-x}

Write the above equation as,

\;\frac{dx}{dy}=e^{-y}-x

\;\frac{dx}{dy}+x=e^{-y}

Compare with

\;\frac{dx}{dy}+Px=Q

So,

P=1, Q=e^{-y}

Therefore,

\begin{aligned} \text{IF} &= e^{\int P\,dy} \\ &= e^{\int 1\,dy} \\ &= e^{y} \end{aligned}

The solution of the given differential equation is,

x ⋅ e^{y} \;\; =∫ e^{y} ⋅ e^{-y} d y+C

x ⋅ e^{y} \;\; =∫ 1 d y+C

x \;\; =e^{-y}(y+C)

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