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  • Solve: (x + y) (dx – dy) = dx + dy. [Hint: Substitute x + y = z after separating dx and dy].

Solve: (x + y) (dx – dy) = dx + dy. [Hint: Substitute x + y = z after separating dx and dy].

 

Answers (1)

Given:

(x+y)(dx-dy)=dx+dy

To find: Solution of the given differential equation

Rewriting the given equation

\left ( x+y \right )\left ( 1-\frac{dy}{dx} \right )=\left ( 1+\frac{dy}{dx} \right )

Assume x+y=z

Differentiate on both sides with respect to x

1+\frac{dy}{dx}=\frac{dz}{dx}

Substituting the values in the equation

z\left ( 1-\frac{dz}{dx}+1 \right )=\frac{dz}{dx}\\ 2z-z\frac{dz}{dx}-\frac{dz}{dx}=0\\ 2z=\left ( z+1 \right )\frac{dz}{dx}\\ dx=\left ( \frac{1}{2}+\frac{1}{2z} \right )dz

Integrate on both the sides

\int dx=\int \left ( \frac{1}{2}+\frac{1}{2z} \right )dz\\ x=\frac{z}{2}+\frac{1}{2}\ln z+c\\ formula:\int \frac{dx}{x}=\ln x
Substitute v=xy

x=\frac{x+y}{2}+\frac{1}{2}\ln(x+y)+\ln c\\ x-y-\ln(x+y)-\ln c=0\\ \ln(x+y)+\ln c=x-y\\ \ln c(x+y)=x-y\\ c(x+y)=e^{x-y}\\ x+y=1/c(e^{x-y})\\ x+y=d(e^{x-y})\\ where d=\frac{1}{c}\\

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