1st order nonlinear homogeneous ODE and integral equations

Question

I'm currently studying ODEs, I've been using a book about the topic, and for now I've seen only first order ODEs and a few methods to solve them like the integrating factor, the Bernoulli, and the two other methods that concerns exact and separable equations.

Then my book gave this exercice under the chapter Miscellaneous Methods of Solving a First Order Differential Equation.

$x^2dy-(xy+y\sqrt{x^2+y^2})dx=0$

I basically used the substitution method for homogeneous equation with $u$ as $u=\frac{y}{x}$ and $dy=udx+xdu$.

I ended up with this: $\int{\frac{\frac{du}{dx}}{u\sqrt{u^2+1}}}dx=ln(\left| x \right|) + C$

My book seemed to assume that I knew how to solve this kind of integral with an unknown function of $x$ or its derivative, I just learnt that it was named an integral equation.

My questions are:

Where can I find informations about how to solve this kind of integrals ? I can't find anything on the Internet about it (is it something we learn later in ODE courses under a different name or something?)

Maybe I should have known how to do this before studying ODEs ? (I'm self thaught, I've only seen Calculus 1 to 3 and integral equation seems to be really physics-related).

Answer 1

Another way:

$$\int \frac{du}{u\sqrt{u^2+1}} $$

Substitute $u=\tan t \implies du=\sec^2 t dt $: $$\int \frac{\sec^2 t}{\sec t \tan t} dt \\ =\int \csc t \ dt \\ = \ln |\csc t -\cot t| + c \\ = \ln \left | \frac{\sqrt{1+u^2}}{u} -\frac 1u \right| + c$$ And now after replacing $u$ by $\frac yx$, you’ll get your solution.

Answer 2

$$I=\int{\frac{\frac{du}{dx}}{u\sqrt{u^2+1}}}dx$$ $$I=\int{\frac{{du}}{u\sqrt{u^2+1}}}$$ $$I=\int{\frac{{udu}}{u^2\sqrt{u^2+1}}}$$ Substitute $u^2+1=v^2 \implies udu=vdv$ then use fraction decomposition. $$I=\int \frac{{dv}}{v^2-1}$$

Answer 3

Note that in this ODE we have

$$ A(x,y)dy-B(x,y)dx=0 $$

and making $y = \lambda x$

$$ \cases{ A(x,y) = x^2A(1,\lambda)\\ B(x,y) = x^2B(1,\lambda)\\ dy = \lambda dx + x d\lambda} $$

$(A,B)$ are same order homogeneous functions: then

$$ \frac{dy}{dx} = \lambda+x\frac{d\lambda}{dx} = \frac{A(1,\lambda)}{B(1,\lambda)} $$

and finally

$$ \frac{dx}{x}=\frac{d\lambda}{\frac{A(1,\lambda)}{B(1,\lambda)}-\lambda} $$