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(x+y+1)dy/dx=1 differential equation ​

Answer»

ANSWER:

\sf <klux>X</klux>+y=Ce^y-2

Step-by-step explanation:

Given:

The differential equation (x + y + 1) dy/dx = 1

To Find:

Solution for the given differential equation

Solution:

\sf (x + y+1)\dfrac{dy}{dx} =1

\sf \dfrac{dy}{dx} =\dfrac{1}{x+y+1}

\sf \dfrac{dx}{dy} =x+y+1

\sf \dfrac{dx}{dy} -x=y+1

Now this is in the form of a LINEAR differential equation,

\sf \dfrac{dx}{dy} +Px=Q

where P = -1 and Q = y + 1

Now finding the INTEGRATING factor (I.F),

\displaystyle \sf I.F=e^{\int\limits {P} \, dy }

Substitute the data,

\displaystyle \sf I.F=e^{\int\limits {-1} \, dy }

\implies \sf e^{-y}

Now the solution of a linear differential equation is given by,

\displaystyle \sf x\times (I.F)=\int\limits {(Q\times I.F)} \, dy +C

Substituting it we get,

\displaystyle \sf x\times e^{-y}=\int\limits {(y+1)\times e^{-y}} \, dy

Now integrating by parts,

\displaystyle \sf x\times e^{-y}=(y+1)\times -e^{-y}-\int\limits {1\times -e^{-y}} \, dy

\displaystyle \sf xe^{-y}=-(y+1)\times e^{-y}-e^{-y}+C

TAKING \sf e^{-y} common,

\sf xe^{-y}=e^{-y}(-y-1-1)+C

Now dividing by \sf e^{-y} we get,

\sf x=-y-2+Ce^y

\sf x+y=Ce^y-2

This is the solution of the given differential equation.


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