# What is a solution to the differential equation (1+y)dy/dx-4x=0?

Jul 29, 2017

The solution is $y = \pm \sqrt{2 {x}^{2} + 1 + {C}_{1}} - 1$

#### Explanation:

$\left(1 + y\right) \frac{\mathrm{dy}}{\mathrm{dx}} - 4 x = 0$

This is a first order ordinary differential equation of the form

$N \left(y\right) \mathrm{dy} = M \left(x\right) \mathrm{dx}$

Therefore,

$\left(1 + y\right) \frac{\mathrm{dy}}{\mathrm{dx}} = 4 x$

$\left(1 + y\right) \mathrm{dy} = 4 x \mathrm{dx}$

Integrating both sides

$\int \left(1 + y\right) \mathrm{dy} = \int 4 x \mathrm{dx}$

$y + {y}^{2} / 2 = 4 {x}^{2} / 2 + C$

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

${y}^{2} + 2 y = 2 {x}^{2} + {C}_{1}$

${y}^{2} + 2 y + 1 = 2 {x}^{2} + 1 + {C}_{1}$

${\left(y + 1\right)}^{2} = 2 {x}^{2} + 1 + {C}_{1}$

$\left(y + 1\right) = \pm \sqrt{2 {x}^{2} + 1 + {C}_{1}}$

$y = \pm \sqrt{2 {x}^{2} + 1 + {C}_{1}} - 1$

Jul 29, 2017

See below.

#### Explanation:

$\frac{1}{2} \left(\frac{d}{\mathrm{dx}}\right) {\left(1 + y\right)}^{2} - 2 \left(\frac{d}{\mathrm{dx}}\right) {x}^{2} = \left(\frac{d}{\mathrm{dx}}\right) \left(\frac{1}{2} {\left(1 + y\right)}^{2} - 2 {x}^{2}\right) = 0$

then

$\frac{1}{2} {\left(1 + y\right)}^{2} - 2 {x}^{2} = C$ or

${\left(1 + y\right)}^{2} = 4 {x}^{2} + {C}_{1}$ or

$1 + y = \pm \sqrt{4 {x}^{2} + {C}_{1}}$ or

$y = - 1 \pm \sqrt{4 {x}^{2} + {C}_{1}}$