xy(dy)/(dx)+4x^2+y^2=0 — step-by-step solution

⚠ 4x is read as the product 4*x: for a power, write 4^x

⚠ xy is read as the product x*y: for a power, write x^y; for the variable xy, write xy_

⚠ xy(dy)/(dx)+4x^2+y^2=0 is read as xy*y'+4x^2+y^2=0: use dif(expression, variable) for a derivative with respect to that variable, use (numerator)/(denominator) for a quotient

Let

z = y^{2}

z' = 2 y \cdot y'

. On intervals where

x

≠ 0, the equation becomes

z' + \frac{2}{x} \cdot z = - 8 x

The integrating factor is

x^{2}

, nonzero for

x

≠ 0: the product rule gives

{(x^{2} z)}^{'}

- x^{3} \cdot 8

⇥Let's compute the integral:

\int(- x^{3} \cdot 8)dx

⇥

- 8 \cdot \frac{x^{4}}{4}

⇥ℹ

- 8 \cdot \frac{x^{4}}{4}

- \frac{8 x^{4}}{4}

‖

- \frac{8 x^{4}}{4}

- 2 x^{4}

⇥

- 2 x^{4}

Integrating gives

x^{2} z

- 2 x^{4}

C

y^{2} = \frac{- 2 x^{4} + C}{x^{2}}

, where

C

is an arbitrary real constant

The real branches have a fixed sign on each interval where

\frac{- 2 x^{4} + C}{x^{2}}

> 0 and

x

≠ 0; these restrictions exclude singular points

▶Solution:

y = {- \sqrt{\frac{- 2 x^{4} + C}{x^{2}}}, \sqrt{\frac{- 2 x^{4} + C}{x^{2}}}}