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  • Find the general solution. ( x + y) dy/dx = 1

Find the general solution.

    Q10.    (x+y)\frac{dy}{dx} = 1

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best_answer

Given equation is
(x+y)\frac{dy}{dx} = 1
we can rewrite it as
\frac{dy}{dx} = \frac{1}{x+y}\\ \\ x+ y =\frac{dx}{dy}\\ \\ \frac{dx}{dy}-x=y
This is  \frac{dx}{dy} + px = Q  type where p =-1 and Q =y
Now,
I.F. = e^{\int pdy}= e^{\int -1 dy}= e^{-y}                     
Now, the solution of given differential equation is given by relation
x(I.F.) =\int (Q\times I.F.)dy +C
x(e^{-y}) =\int y\times e^{-y}dy +C
xe^{-y}= \int y.e^{-y}dy + C
Lets take
 I=\int ye^{-y}dy \\ \\ I = y .\int e^{-y}dy-\int \left ( \frac{d(y)}{dy}.\int e^{-y}dy \right )dy\\ \\ I =- y.e^{-y}+ \int e^{-y}dy\\ \\ I = - ye^{-y}-e^{-y}
Put this value in our equation
x.e^{-y} = -e^{-y}(y+1)+C\\ x = -(y+1)+Ce^{y}\\ x+y+1=Ce^y
Therefore, the general solution is   x+y+1=Ce^y

Posted by

Gautam harsolia

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