Question

Solve the given differential equations $xy{p}^2+(x^2+xy+y^2)p+x^2+xy=0$ where $p=\frac{dy}{dx}$

• A. $(xy+x^2-C)(x^2+y^2-C)=0$
• B. $(2xy+x^2-C)(x^2+y^2-C)=0$
• C. $(2xy+x^2-C)(x^3+y^3-C)=0$
• D. $(2xy+x^3-C)(x^2+y^2-C)=0$

Answer 1

The correct option is B $(2xy+x^2-C)(x^2+y^2-C)=0$

the giving differential equation is given by
$xy{p}^2+(x^2+xy+y^2)p+x^2+xy=0$
The above equation can be factored into $(xp+y+x)(yp+x)=0$
Now, the solution of the component equation will be
$x\frac{dy}{dx}+x+y=0$ and $y\frac{dy}{dx}+x=0$
$ydy=-xdx$
Integrate both the sides, we get
$\frac{y^2}{2}=\frac{-x^2}{2}+c_2$
$x^2+y^2+c=0$
A first order linear Ordinary differential equation is given by $y^{\prime }+p\left(x\right)y=q\left(x\right)$
The solution of the ODE will be $y\left(x\right)=\frac{\int e^{\int p\left(x\right)dx}q\left(x\right)dx+C}{e^{\int p\left(x\right)dx}}$
Here, $p\left(x\right)=\frac{1}{x}$ and $q\left(x\right)=-1$
Here, intergating factor will be $\mu =e^{\int \frac{1}{x}dx}=x$
So,
$y=\frac{\int -xdx+c_1}{x}=\frac{-x^2+c}{2x}$
$2xy+x^2-c=0$
Hence the primitive will be $(2xy+x^2-c)(x^2+y^2-c)=0$