Question

Use the substitution to reduce the equation $y^3\frac{dy}{dx}+x+y^2=0$ to homogeneous form and hence solve it. The solution is

$\frac{1}{2}\ln |x^2+a^2|-{\tan }^{-1}\left(\frac{a}{x}\right)=c$, then $a=?$

• A. $a=x+y^2$
• B. $a=x-y^2$
• C. $a=\sqrt{(}x+y^2)$
• D. $a=x+y^3$

Answer 1

The correct option is A $a=x+y^2$

Put $y^2=v-x\Rightarrow y\frac{dy}{dx}=\frac{1}{2}(\frac{dy}{dx}-1)$
The given equation is
$y^2.\left(y\frac{dy}{dx}\right)+x+y^2=0\Rightarrow (v-x).\frac{1}{2}(\frac{dv}{dx}-1)+x(v-x)=0$
$\Rightarrow (v-x)\frac{dv}{dx}-(v-x)+2v=0\Rightarrow \frac{dv}{dx}=\frac{v+x}{x-v}$ ...(1)
Thus the given equation reduced to homogeneous form,
Now to solve it put $v=zx$
So that $z+x\frac{dz}{dx}=\frac{z+1}{1-z}$
Therefore from (1) we get
$z+x\frac{dz}{dx}=\frac{z+1}{1-z}\Rightarrow x\frac{dz}{dx}=\frac{z+1}{1-z}-z=\frac{1+z^2}{1-z}$
$\Rightarrow \frac{(1-z)dz}{1+z^2}=\frac{dx}{x}\Rightarrow \frac{1}{1+z^2}dz-\frac{zdz}{1+z^2}=\frac{dx}{x}$
Integrating, we get
${\tan }^{-1}z-\frac{1}{2}\log (1+z^2)=\log x+\log c$
$\Rightarrow {\tan }^{-1}\left(\frac{y^2+x}{x}\right)-\frac{1}{2}\log (x^2+{(y^2+x)}^2)=\log c$