Answer to Question #116885 in Differential Equations for Vikas

Question

identify the type of the differential equation y =xdy/dx+1-logdy/dx and hance solve it

Answer 1

We have given the differential equation as

$$y =x\frac{dy}{dx}+1-log\big(\frac{dy}{dx}\big) \hspace{1cm}(\clubs)$$

Clearly, it has the form of

$$y =x\big(\frac{dy}{dx}\big)+1-H\big(\frac{dy}{dx}\big)$$

i.e

$$H=H(\frac{dy}{dx})$$

Thus, this form of the differential equation is called Clairaut's equation.

Let's fix the notation for

$$y'=\frac{dy}{dx} \& \: y''=\frac{d^2y}{dx^2}$$

Now, to find the solution of the equation $(\clubs)$ ,differentiate both side of the equation with respect to $x$ ,hence we get,

$$y'=xy''+y'-\bigg(\frac{1}{y'}y''\bigg)\\ \implies \big(x-\frac{1}{y'}\big)y''=0 \hspace{1cm}(\spades)$$

Clearly, equation $(\spades)$ has two cases.

Case-I:

Suppose

$$y''=0$$

we already know the solution(which is trivial) i.e

$$y'=a \\ \implies y=ax+b$$

where,$a,b$ are any constant value.

Now, if we plugin the above solution to $(\clubs)$ we get,

$$y=ax+1-log(a) \iff y=ax+b$$

Case-II:

Suppose

$$x-\frac{1}{y'}=0\\ \implies y'=\frac{1}{x}\\ \implies \int dy=\int\frac{1}{x}dx\\ \implies y=log(x)+c$$

If we plugin the above solution to $(\clubs)$ then we get,

$$c=2$$

Hence, the final solution will be in the form of

$$y=log(x)+2$$

Interestingly, this solution is singular!

Hence, we are done.