Question

Find a particular solution of the following differential equation:- x² dy + (xy + y²) dx = 0; y = 1 when x = 1

Answer 1

We have,

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

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

Let y = vx

$$\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}$$

$$\therefore v+x\frac{dv}{dx}=-\left(\frac{v{x}^2+v^2{x}^2}{x^2}\right)$$

$$\Rightarrow x\frac{dv}{dx}=-(v+v^2)-v$$

$$\Rightarrow x\frac{dv}{dx}=-2v-v^2$$

$$\Rightarrow \frac{1}{v^2+2v}dv=-\frac{1}{x}dx$$

Integrating both sides, we get

$$\int \frac{1}{v^2+2v}dy=-\int \frac{1}{x}dx$$

$$\int \frac{1}{v^2+2v+1-1}dy=-\int \frac{1}{x}dx$$

$$\int \frac{1}{{(v+1)}^2-{\left(1\right)}^2}dy=-\int \frac{1}{x}dx$$

$$\Rightarrow \frac{1}{2\times 1}\log \left|\frac{v+1-1}{v+1+1}\right|=-\log \left|x\right|+\log C$$

$$\Rightarrow \frac{1}{2}\log \left|\frac{v}{v+2}\right|=-\log \left|x\right|+\log C$$

$$\Rightarrow \log \left|\frac{v}{v+2}\right|=-2\log \left|x\right|+2\log C$$

$$\Rightarrow \log \left|\frac{v}{v+2}\right|+\log \left|x^2\right|=\log C^2$$

$$\Rightarrow \log \left|\frac{v{x}^2}{v+2}\right|=\log C^2$$

$$\Rightarrow \left|\frac{v{x}^2}{v+2}\right|=C^2$$

$$\Rightarrow \left|\frac{v{x}^2}{v+2}\right|=k,\text{ where }k=2C$$

$$\Rightarrow \left|\frac{\frac{y}{x}{x}^2}{\frac{y}{x}+2}\right|=k$$

$$\Rightarrow \left|\frac{x^{2}y}{y+2x}\right|=k.....\left(1\right)$$

Now,

When x = 1, y = 1

$$\therefore \left|\frac{1}{1+2}\right|=k$$

$$\Rightarrow k=\frac{1}{3}$$

Putting the value of k in (1), we get

$$\left|\frac{x^{2}y}{y+2x}\right|=\frac{1}{3}$$

$$\Rightarrow 3x^{2}y=\pm (y+2x)$$

$$\text{But }y\left(1\right)=1\text{ does not satisfy the equation }3x^{2}y=-(y+2x)$$

$$\therefore 3x^{2}y=y+2x$$

$$\Rightarrow y=\frac{2x}{3x^2-1}$$