Question

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

Answer 1

Answer:

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

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.