The solution of the equation dy/dx=1/x+y+1 is

Correct Answer is : log⁡(x+y+2)=C+y (x+y+2) = C + ylog(x+y+2)=C+y

Correct Answer is : log⁡(x+y+2)=C+y (x+y+2) = C + ylog(x+y+2)=C+y

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CGPET 2015 Solved Paper

Section: Mathematics

Question : 102 of 150

Marks: +1, -0

\frac{dy}{dx}=\frac{1}{x+y+1}

Solution:

Given differential equation is

\frac{dy}{dx} = \frac{1}{x+y+1}

\frac{dx}{dy} = x+y+1 \Rightarrow \frac{dx}{dy} - x = y+1

...(i)

Eq. (i) is of the type

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

, where P and Q

are functions of y or constant terms.

Here,

P=-1

and

Q = y=1

∴

IF = e^{\int P\,dy} = e^{\int (-1)\,dy} = e^{-y}

Now, general solution is given by

x \cdot IF = \int IF \cdot Q\,dy + C_1

⇒

x e^{-y} = \int e^{-y}(y+1)\,dy + C_1

⇒

x e^{-y} = (y+1)\left(\frac{e^{-y}}{-1}\right) + \int 1\cdot e^{-y}\,dy + C_1

⇒

x e^{-y} = -e^{-y}(y+1) - e^{-y} + C_1

⇒

x = -(y+1)-1+Ce^y \text{ [on dividing by } e^{-y}\text{]}

⇒

x+y+2 = Ce^y

On taking log both sides, we get

log

(x+y+2) = \log C + 1 e^y

⇒ log

(x+y+2) = \log C_1 + \log e^y

[ \because \log mn = \log m + \log n ]

[ \text{put } C = \log C_1 ]

]

⇒log

(x+y+2) = C + y

which is the required solution.

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