Math / Calculus

QuestionClairaut requation $y=4 x y^{\prime}+\ln y^{\prime}$

Studdy Solution

STEP 1

1. The given equation is a Clairaut equation, which is a type of differential equation.
2. Clairaut's equations are of the form $y = xy' + f(y')$ , where $f$ is a function of $y'$ .
3. The solution involves finding both the general solution and the singular solution.

STEP 2

1. Identify the form of the Clairaut equation.
2. Differentiate the equation with respect to $x$ .
3. Solve for $y'$ in terms of $x$ .
4. Find the general solution.
5. Determine the singular solution.

STEP 3

Identify the form of the Clairaut equation. The given equation is:
$y = 4xy' + \ln y'$
This matches the form $y = xy' + f(y')$ with $f(y') = 4xy' + \ln y'$ .

STEP 4

Differentiate the equation with respect to $x$ :
$\frac{d}{dx}(y) = \frac{d}{dx}(4xy' + \ln y')$
Using the product rule and chain rule:
$y' = 4(y' + x \frac{d}{dx}(y')) + \frac{1}{y'} \frac{d}{dx}(y')$

STEP 5

Simplify the differentiated equation:
$y' = 4y' + 4x y'' + \frac{y''}{y'}$
Rearrange to solve for $y''$ :
$0 = 4x y'' + \frac{y''}{y'}$
Factor out $y''$ :
$y''(4x + \frac{1}{y'}) = 0$

STEP 6

Solve for $y'$ in terms of $x$ :
Since $y'' = 0$ , we have $y' = c$ , where $c$ is a constant.

STEP 7

Find the general solution by substituting $y' = c$ back into the original equation:
$y = 4xc + \ln c$
This represents a family of straight lines parameterized by $c$ .

STEP 8

Determine the singular solution by considering the derivative $\frac{d}{dc}(4xc + \ln c) = 0$ :
$4x + \frac{1}{c} = 0$
Solve for $c$ :
$c = -\frac{1}{4x}$
Substitute back to find $y$ :
$y = 4x(-\frac{1}{4x}) + \ln(-\frac{1}{4x})$
$y = -1 + \ln(-\frac{1}{4x})$
This is the singular solution.
The general solution is $y = 4xc + \ln c$ and the singular solution is $y = -1 + \ln(-\frac{1}{4x})$ .

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