Mathematics - Differential Equation Question with Solution | TestHub

We are given the differential equation:

$$\frac{xdx+ydy}{ydx-xdy}=\frac{x^4+2x^2{y}^2+y^4}{x^2}$$

First, we can simplify the numerator on the right side: $x^4+2x^2{y}^2+y^4=(x^2+y^2{)}^2$.

So the equation becomes:

$$\frac{xdx+ydy}{ydx-xdy}=\frac{(x^2+y^2{)}^2}{x^2}$$

Rearranging the terms, we get:

$$\frac{xdx+ydy}{(x^2+y^2{)}^2}=\frac{ydx-xdy}{x^2}$$

Now, let's recognize the forms of the differentials.

The left side can be integrated using the substitution $u=x^2+y^2$, so $du=2xdx+2ydy$.

Thus, $\frac{xdx+ydy}{(x^2+y^2{)}^2}=\frac{1}{2}\frac{2xdx+2ydy}{(x^2+y^2{)}^2}=\frac{1}{2}d\left(\frac{-1}{x^2+y^2}\right)$.

The right side is related to the differential of $\frac{y}{x}$:

$$d\left(\frac{y}{x}\right)=\frac{xdy-ydx}{x^2}=-\frac{ydx-xdy}{x^2}$$

So, $\frac{ydx-xdy}{x^2}=-d\left(\frac{y}{x}\right)$.

Substituting these back into the rearranged equation:

$$\frac{1}{2}d\left(\frac{-1}{x^2+y^2}\right)=-d\left(\frac{y}{x}\right)$$

Integrating both sides:

$$\int \frac{1}{2}d\left(\frac{-1}{x^2+y^2}\right)=\int -d\left(\frac{y}{x}\right)$$

$$-\frac{1}{2(x^2+y^2)}=-\frac{y}{x}+C^{\prime }$$

Multiplying by 2 and rearranging to solve for the constant $C$:

$$\frac{-1}{x^2+y^2}=-2\frac{y}{x}+2C^{\prime }$$

$$2\frac{y}{x}-\frac{1}{x^2+y^2}=2C^{\prime }$$

Let $C=2C^{\prime }$. The final solution is:

$$2\frac{y}{x}-\frac{1}{x^2+y^2}=C$$