What is the solution for x+y-1+(x-y+3)y^'=0 ?

The solution for x+y-1+(x-y+3)y^'=0 is y=3+x-sqrt(2x^2+4x+9-c_{1)},y=x+sqrt(2x^2+4x+9-c_{1)}+3

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x+y-1+(x-y+3)y^'=0

x + y - 1 + (x - y + 3) y^{^{'}} = 0

y = 3 + x - 2 x^{2} + 4 x + 9 - c_{1}, y = x + 2 x^{2} + 4 x + 9 - c_{1} + 3

Solution steps

x + y - 1 + (x - y + 3) y^{'} (x) = 0

Solve exact ODE:

y = 3 + x - 2 x^{2} + 4 x + 9 - c_{1}, y = x + 2 x^{2} + 4 x + 9 - c_{1} + 3

y = 3 + x - 2 x^{2} + 4 x + 9 - c_{1}, y = x + 2 x^{2} + 4 x + 9 - c_{1} + 3

5 x^{2} e^{x}

\int_{0}^{1} (2 x + 3) e^{- 5 x} d x

\frac{d}{d x} (\frac{e ^{x} ( x ^{5} + 3 )}{x ^{10} + 9})

\frac{s}{s ^{2} + a ^{2}}

\int \frac{6 x ^{3} + 4 x ^{2} - 47 x - 27}{x ^{2} - 9} d x

What is the solution for x+y-1+(x-y+3)y^'=0 ?
The solution for x+y-1+(x-y+3)y^'=0 is y=3+x-sqrt(2x^2+4x+9-c_{1)},y=x+sqrt(2x^2+4x+9-c_{1)}+3