Question

Identifying the differential equation and solve them
y=xy' + 1 - lny'

Answer 1

Answer on Question #75288- Math - Differential Equations

Identifying the differential equation and solve them $y = xy' + 1 - lny'$.

Solution

This equation of a look:

$$y(x) = x \frac{dy}{dx} + f\left(\frac{dy}{dx}\right)$$

is called Clairaut's equation.

Use the next replacement:

$$\frac{dy}{dx} = p,$$

then

$$y = xp + 1 - \ln p$$

Let's differentiate

$$\frac{dy}{dx} = p + x \frac{dp}{dx} - \frac{1}{p} \cdot \frac{dp}{dx},$$

then

$$\begin{array}{l} p = p + x \frac{dp}{dx} - \frac{1}{p} \cdot \frac{dp}{dx} \\ x \frac{dp}{dx} - \frac{1}{p} \cdot \frac{dp}{dx} = 0 \\ \frac{dp}{dx}\left(x - \frac{1}{p}\right) = 0 \\ \end{array}$$

Equate each multiplier to zero:

$$\begin{array}{l} \frac{dp}{dx} = 0 ; \\ x - \frac{1}{p} = 0 . \end{array}$$

Integrate equality (2):

$$p = C,$$

let's substitute this value in (1): $y = Cx + 1 - \ln C$ - common decision.

From (3):

$$x - \frac{1}{p} = 0, x = \frac{1}{p}, p = \frac{1}{x}$$

let's substitute this value in (1):

$$\begin{array}{l} y = x \frac{1}{x} + 1 - \ln \frac{1}{x} \\ y = 1 + 1 - \ln x^{-1} \\ y = 2 + \ln x - \text{singular solution} \\ \end{array}$$

Answer: $y = Cx + 1 - \ln C, y = 2 + \ln x$.

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