Clairauts Equation and Its General and Singular Solutions

The equation of the clairaut's form is as given: y = xy′+ ψ(y′), where ψ(y′) = a nonlinear differentiable function. The Clairauts equation is a special case of the Lagrange equation when φ(y′) equals y′. It is solved in the same manner by establishing a parameter. A standard solution is assigned as y = Cx + ψ(C), where C = an arbitrary constant.

Likewise, the Lagrange equation, the Clairaut equation also may contain a singular solution which is expressed parametrically in the form:

x = −ψ′(p)

y = xp + ψ(p)

,where p = parameter.

Clairaut Differential Equation

Clairaut form of the differential equation is basically a 1st -order differential equation of the form: y = x dy/dx + f (dy/dx)

In which, f is a suitable function.

Differential Equation Solution Method and Formula

Differentiate both sides in terms of x and obtain:

dy/dx = dy/dx + (x + f' (dy/dx)) d²y/d²x

Cancel the common term from both the sides of equation and get:

0 = (x + f' (dy/dx)) d²y/d²x

This provides two possible solution types as below:

D²y/d²x = 0

x + f'(dy/dx) = 0

Remember that when we differentiate, we lose details, so it is not true that all solutions of the differentiated equation solve the actual equation. Instead, we require plugging these solutions into the actual equation to compel them. We get the following:

• The solutions for the d²y/d²x = 0 case are straight lines in the form y = Cx + f(C). This is the solution family for the general solution of the differentiated equation.

• The solution for the x + f'(dy/dx) = 0 case is special: it is a parametric curve provided by x = -f'(p), y = f(p) - pf'(p) in which p = dy/dx is the parameter moving along the curve. This is referred to as the singular solution. The curve resembling this is the envelope of the general solution curves. For the purpose of verifying that this solution is right, we can compute dy/dx with the help of parametric differentiation with reference to p and check that it is indeed equivalent to p.

Solved Example

Example:

1. Evaluate the general and singular solutions of the clairaut differential equation form y = 2xy' – 3 (y') 2.

Solution:

Here we observe that we manage a Lagrange equation. We will solve it with the help of the method of differentiation.

Represent y′= p, so the equation is expressed in the form:

y

Differentiating both sides of the equation, we find:

dy = 2xdp + 2pdx − 6pdp.

Now, you can replace dy with pdx:

pdx = 2xdp + 2pdx − 6pdp, ⇒ −pdx = 2xdp − 6pdp.

Dividing by p, we can express the following mathematical equation (further we check if p = 0 is a solution of the original equation):

−dx = 2xpdp − 6dp, ⇒ dxdp + 2px – 6 = 0.

As it can be observed, we get a linear equation for the function x (p). The integrating factor is \[u(p) = exp (\int \frac{2p}{dp}) = exp(2ln\left| p\right|) = exp(ln\left| p\right|2) = |p^{2} = p^{2}\].

The general solution of the linear equation of the form is assigned as:

\[X(p) = \int p^{2}.6dp + \frac{C}{p^{2}} = \frac{6p^{3}}{3} + \frac{C}{p^{2}} = 2p + \frac{C}{p^{2}}\].

By substituting this expression for x into the Lagrange equation, we get:

y = 2(2p + Cp²)p − 3p² = 4p² + 2Cp − 3p² = p² + 2Cp.

Therefore, the general solution in parametric form is explained by the methodology of equations:

{X (p) = 2p + C/p²

{X (p) = p² + 2C/p

Apart from that, the Lagrange equation can contain a singular solution. Solving the equation φ (p) − p = 0, we determine the root:

2p − p = 0, ⇒p = 0.

Thus, the singular solution is written by the linear function:

y=φ (0)x + ψ(0) = 0⋅x + 0 = 0.