It is my first week dealing with Differential Equations, and I am stuck at the following question:

Find all the solutions of the following equation:

$2y^2dx-(x+y)^2dy=0$

I have consulted an online calculator, which gave the solution $log\frac{y(x)}{x}+2arctan\frac{y}{x}=c_1-log(x)$.

However, I would like to be able to do this manually.

Could anybody help out?

Hint:

$$\dfrac{dy}{dx}=\dfrac{2y^2}{(x+y)^2}=\dfrac{2\left(\dfrac yx\right)^2}{\left(1+\dfrac yx\right)^2}$$

As the denominator & numerator are Homogeneous polynomials

Set $y=vx,\dfrac{dy}{dx}=v+x\dfrac{dv}{dx}$

July 12, 2019 09:33 AM

The differential equation

$$2y^2dx-(x+y)^2dy=0$$

can be rewriten as

$$2\dfrac{dx}{dy}=\left(\dfrac{x+y}{y}\right)^2$$

or

$$2\dfrac{dx}{dy}=\left(\dfrac{x}{y}+1\right)^2\ \ \ ...(1)$$

Now substituting $\dfrac{x}{y}$ as t, or $x=yt$

If you differentiate $x=yt$ w.r.t. $y$ on both sides you'll get $\dfrac{dx}{dy}=t+y\dfrac{dt}{dy}$

Substituting the value of $\dfrac{dx}{dy} \text{ and } \dfrac{x}{y}$ in equation (1)

$$2t+2y\dfrac{dt}{dy}=(t+1)^2$$

$$2t+2y\dfrac{dt}{dy}=t^2+1+2t$$

$$2y\dfrac{dt}{dy}=t^2+1$$

$$2\dfrac{dt}{t^2+1}=\dfrac{dy}{y}$$

Now integrating both sides

$$\displaystyle \int 2\dfrac{dt}{t^2+1}=\int \dfrac{dy}{y}$$

$$2\tan ^{-1} t=\ln(y)+C$$

Substituting the value of t in this expression

$$2\tan ^{-1} \dfrac{x}{y}=\ln(y)+C$$

In this question integrating $2\dfrac{dx}{dy}=\left(\dfrac{x+y}{y}\right)^2$ was much easier than integrating $$$

Although my solution doesn't match yours it doesn't mean its incorrect, its a matter of integration constant.

July 12, 2019 10:11 AM

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