Solve Clairaut's Equation: Find Singular Solution

[Suvadip]

Please help me to find the singular solution of the following Clairaut's equation

\(\displaystyle y=px+\sqrt{1+2p^4}\) where \(\displaystyle p=\frac{dy}{dx}\)

 

[MarkFL]

What progress have you made?

Here are some links to get you started:

Clairaut's Differential Equation -- from Wolfram MathWorld

Clairaut's equation - Wikipedia, the free encyclopedia

 

[Suvadip]

I can express as x=f(p) and y=g(p). But I can't eliminate p to find a relation between x and y.

What progress have you made?

Here are some links to get you started:

Clairaut's Differential Equation -- from Wolfram MathWorld

Clairaut's equation - Wikipedia, the free encyclopedia

 

[Fernando Revilla]

I can express as x=f(p) and y=g(p). But I can't eliminate p to find a relation between x and y.

Is it necessary to express the singular solution in implicit form?. You can write it in parametric form: $$x=\dfrac{-4p^3}{\sqrt{1+2p^4}},\quad y=\dfrac{1-2p^4}{\sqrt{1+2p^4}}$$

 

[blue_raver22]

To solve Clairaut's equation and find the singular solution, we need to first differentiate the given equation with respect to x. This gives us:

p = p + x(p) + \frac{2p^3}{\sqrt{1+2p^4}}\frac{dp}{dx}

Next, we can simplify this equation by rearranging terms and isolating the derivative term:

\frac{dp}{dx} = \frac{\sqrt{1+2p^4}}{2p^3}

Now, we can substitute this expression for the derivative back into the original equation and solve for p:

y = px + \sqrt{1+2p^4}
y = px + \frac{2p^3}{2p^3}
y = px + 1

This is the singular solution of Clairaut's equation, as it only contains one arbitrary constant (p) and does not involve differentiation. This solution represents a family of curves that all satisfy the original equation and have the same slope at every point.

In conclusion, the singular solution of Clairaut's equation is y = px + 1.