xydx-(1+x^2)dy=0 — step-by-step solution

⚠ xydx-(1+x^2)dy = 0 is read as the differential equation y' = (xy)/(1+x^2): divide by dx and isolate y', the derivative of y with respect to x■xy is read as the product x*y: for the variable xy, write xy_

1) Solve the separable equation

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

, where

y

is a function of

x

Before dividing,

y = 0

gives the equilibrium solution

y = 0

On intervals where

y \neq 0

, separate the variables:

\frac{dy}{y} = \frac{x}{x^{2} + 1} \cdot dx

⇥1.1) Let's compute the integral:

\int \frac{1}{y} d y

⇥ⓘ The derivative of

\ln | y |

\frac{1}{y}

, so

\ln | y |

is an antiderivative of

\frac{1}{y}

\int \frac{1}{y} d y

\ln | y |

⇥

\ln | y |

⇥1.2) Let's integrate:

\int \frac{x}{x^{2} + 1} d x

⇥For the math expression:

\frac{x}{x^{2} + 1}

, since (ln(a))' = a'/a, we try differentiating

\ln (x^{2} + 1)

⇥1.3) Let's differentiate:

\frac{d}{d x} (\ln (x^{2} + 1))

⇥We differentiate using:

\frac{d}{d x} (\ln (f)) = \frac{d}{d x} (f) \cdot \frac{1}{f}

(where f =

x^{2} + 1

)

⇥ⓘ The derivative of ln(x) is 1/x:

\frac{d}{d x} (\ln (x^{2} + 1))

\frac{d}{d x} (x^{2} + 1) \cdot \frac{1}{x^{2} + 1}

⇥

\frac{d}{d x} (x^{2} + 1) \cdot \frac{1}{x^{2} + 1}

⇥ⓘ Multiplying a fraction by a term multiplies its numerator:

\frac{d}{d x} (x^{2} + 1) \cdot \frac{1}{x^{2} + 1}

\frac{\frac{d}{d x} (x^{2} + 1) \cdot 1}{x^{2} + 1}

\frac{\frac{d}{d x} (x^{2} + 1)}{x^{2} + 1}

■The derivative of a sum is the sum of the derivatives:

\frac{d}{d x} (x^{2} + 1)

\frac{d}{d x} (x^{2}) + 0

⇥

\frac{\frac{d}{d x} (x^{2}) + 0}{x^{2} + 1}

⇥ⓘ Adding 0 changes nothing:

\frac{d}{d x} (x^{2}) + 0

\frac{d}{d x} (x^{2})

■The derivative of x^n is n*x^(n-1):

\frac{d}{d x} (x^{2})

2 x

⇥

\frac{2 x}{x^{2} + 1}

⇥You observe that the computed derivative and the original math expression are the same, except for a coefficient. So the integral is:

⇥ⓘ The derivative of

\frac{1}{2} \cdot \ln (x^{2} + 1)

\frac{x}{x^{2} + 1}

, so

\frac{1}{2} \cdot \ln (x^{2} + 1)

is an antiderivative of

\frac{x}{x^{2} + 1}

\int \frac{x}{x^{2} + 1} d x

\frac{1}{2} \cdot \ln (x^{2} + 1)

⇥

\frac{1}{2} \cdot \ln (x^{2} + 1)

⇥ⓘ Multiplying a fraction by a term multiplies its numerator:

\frac{1}{2} \cdot \ln (x^{2} + 1)

\frac{1 \cdot \ln (x^{2} + 1)}{2}

\frac{\ln (x^{2} + 1)}{2}

⇥

\frac{\ln (x^{2} + 1)}{2}

Integrating both sides gives

\ln | y | = \frac{\ln (x^{2} + 1)}{2} + C

, where

C

is an arbitrary real constant

Taking the exponential gives

| y | = \exp (C) \exp (\frac{\ln (x^{2} + 1)}{2})

\exp (C) \cdot \sqrt{x^{2} + 1}

, hence

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

with

K

= ±

\exp (C)

, a nonzero real constant

K = 0

gives back the equilibrium solution

y = 0

: the solutions are

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

, where

K

is an arbitrary real constant

▶Solution:

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