How to solve Clairaut Type Differential Equations?

In the previous post, we learnt about the basics of Clairaut's Equation. Now, let us see how easy it becomes to solve differential equations of the form

Where p=y' using the following results 

The General Solution

And the SINGULAR SOLUTION (if it exists)

We will also see what is an envelope of a curve.

Let us start by taking some examples. Before we solve the equation in the thumbnail, let us start by some direct and basic examples.

How will you solve? 

Let us put y'=p and re-write the equation

This is of the above mentioned form where

Therefore, the general solution can be found very easily, simply replace p with a constant c, so the GENERAL SOLUTION bceomes 

See how easily we found out the solution to our differential equation. No integration. Nothing. Simply by replacing p with c because it was in the form of a Clairaut's Equation. How easy! 

Now let's find out the SINGULAR SOLUTION 

Differentiating, 

Since f'(p)=-x,

Now if we put the value of p in 

, we get our SINGULAR SOLUTION 

Now this singular solution is actually the envelope of the general solution

Let us understand this through the graph of the general solution and the singular solution

Now if we see the graph, the general solutions are obtained by varying the value of the constant c. Here the general solution is a family of straight lines. Now the parabola you see is the singular solution. As it was said earlier, the singular solution is the envelope of the general solution, that is, the singular solution will be tangent to each and every curve we obtain through the general solution. Now it can be seen very clearly, that no matter what straight line we obtain by putting the value of c, it is always tangent to the parabola, our singular solution. 

Noticed how easily we solved a first order differential equation of degree 2?     

Let's take a look at a similar example

How can we solve the differential equation

Let y'=p and re-write the equation

Now it becomes a Clairaut's Equation with

The GENERAL SOLUTION is simply

And the SINGULAR SOLUTION can be found very easily as

Now we can substitute the value of p in the original differential equation and obtain

Again see the graph

As we change the value of c in general solution, we get different solutions which here is a family of straight lines and notice how the  singular solution remains tangent to every straight line we obtain through the general solution, that is, the singular solution is the envelope of general solution. Again, we solved a third degree first order differential equation so easily as it was a Clairaut Equation.

Well, in these examples I took up, the equations were already in the form of Clairaut's Equation. It is also possible to reduce an equation in the form of Clairaut's Equation and solve it very easily.

Let us take a look at this.

How can we solve 

Let's put y'=p and simplify a bit

Now it doesn't look like a Clairaut Equation

But is it possible to convert it into the form of Clairaut's Equation?

Let us try.

Let us make some substitutions 

And

Differentiating

We get 

And

Dividing equation 1 by equation 2 and simplifying, we get

where P=dv/du

Let us substitute this value of p in the simplified differential equation

Making the substitution and making further simplifications we get 

Now notice this has taken the form of Clairaut Equation, only the x and y terms have been replaced by x² and y². Can we apply the result of Clairaut's Equation in this equation too? Of course we can. Simply replace P with a constant c and we get our general solution.

Here is the graph of the general solution

How simple!
Now I will leave it up to you to find the SINGULAR SOLUTION. Can you tell if it exists? And if it does what does it represent? Remember to drop in your equations of the singular solution in the comments below.

Till next time,

Thanks for reading

And

Happy Learning

Graphs created using Desmos