What is the solution for y^'=(x+y)^2-1,y(1)=6 ?

The solution for y^'=(x+y)^2-1,y(1)=6 is y=-7/(7x-8)-x

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y^'=(x+y)^2-1,y(1)=6

y^{^{'}} = (x + y)^{2} - 1, y (1) = 6

y = - \frac{7}{7 x - 8} - x

Solution steps

y^{'} (x) = (x + y)^{2} - 1

Solve using substitution:

y = - \frac{7}{7 x - 8} - x

y = - \frac{7}{7 x - 8} - x

x \to 1 - lim (x^{3})

sin (7 ln (x))

\int sin^{3} (x) cos^{9} (x) d x

\frac{\partial}{\partial x} (\frac{x}{y - x})

\frac{\partial}{\partial x} (x + yz)

What is the solution for y^'=(x+y)^2-1,y(1)=6 ?
The solution for y^'=(x+y)^2-1,y(1)=6 is y=-7/(7x-8)-x