What is the solution for (x+y)^2dx+(x+y)^2dy=0 ?

The solution for (x+y)^2dx+(x+y)^2dy=0 is y=-x+c_{1}

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(x+y)^2dx+(x+y)^2dy=0

(x + y)^{2} d x + (x + y)^{2} d y = 0

y = - x + c_{1}

Solution steps

(x + y)^{2} d x + (x + y)^{2} d y = 0

Let

y

be the dependent variable. Divide by

d x

(x + y)^{2} + (x + y)^{2} \frac{d y}{d x} = 0

Substitute

\frac{d y}{d x}

with

y^{'} (x)

(x + y)^{2} + (x + y)^{2} y^{^{'}} (x) = 0

Isolate

y^{'} (x) : y^{'} (x) = - 1

f^{^{'}} (x) = g (x)

then

f (x) = \int g (x) d x

y = \int - 1 d x

\int - 1 d x = - x + c_{1}

y = - x + c_{1}

y = \frac{10}{x ^{2} - 4 x - 21}, x = - 2, x = 2

g (x) = x^{3} - 6 x^{2} + 9 x + 3

x = 4 - y^{2}, x = y^{2} - 4

\int \frac{p}{( p - 5 ) ^{2}} d p

\frac{d y}{d x} = 2 y + x^{2} + 9

What is the solution for (x+y)^2dx+(x+y)^2dy=0 ?
The solution for (x+y)^2dx+(x+y)^2dy=0 is y=-x+c_{1}