The solution of dy-dx-xy-x-y-1

The correct option is C y+1=c.e(x^2+2x/2) dydx = x(y+1)+(y+1) ⇒ #160;∫dy(y+1) = ∫ dx (x+1) + c ⇒ Log (y+1) = x^22 + x+c ⇒ (y+1) = c. ...

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Variable Separable Method

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Question

The solution of $\frac{d y}{d x} = x y + x + y + 1$

y = \frac{x^{2}}{2} + x c

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x = \frac{y^{2}}{2} + y + c

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y + 1 = c . e (\frac{x^{2} + 2 x}{2})

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y + 1 = x (x + 1)

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\frac{d y}{d x} = 1 + y + y^{2} + x + x y + x y^{2}

\frac{d y}{d x}

y = \frac{1 + x}{1 - x}

y = \frac{1 - x}{1 + x}

The solution of the differential equation $\frac{d y}{d x} = 1 + x + y + x y$ is

y = e^{x} + 1

y^{''} - y = 0

y = x^{2} + 2 x + C

y^{'} - 2 x - 2 = 0

y = cos x + C

y^{'} + sin x = 0

y = \sqrt{1 + x^{2}}

y^{'} = \frac{x y}{1 + x^{2}}

y = A x

x y = y (x \neq 0)

y = x sin x

x y = y + x \sqrt{x^{2} y^{2}} (x \neq 0 a n d x > y o r x < y)

x y = log y + C

y^{'} = \frac{y^{2}}{1 - x y} (x y \neq 1)

y - cos y = x

(y sin y + cos y + x) y^{'} = y

x + y = {tan}^{- 1} y

y^{2} y^{'} + y^{2} + 1 = 0

y = \sqrt{a^{2} - x^{2}} x ϵ (- a, a)

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

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

y (1) = 1

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