Question

Use the substitution $y^2=a-x$ to reduce the equation $y^3.\frac{dy}{dx}+x+y^2=0$ 0 to homogeneous form and hence solve it. (where a is variable)

• A. $\frac{1}{2}ln|x^2-a^2|-ta{n}^{-1}\left(\frac{a}{y}\right)=c$ where $a=x+y^2$
• B. $\frac{1}{2}ln|x^2+a^2|-co{t}^{-1}\left(\frac{a}{x}\right)=c$ where $a=x+y^2$
• C. $\frac{1}{2}ln|x^2-a^2|-co{t}^{-1}\left(\frac{a}{y}\right)=c$ where $a=x+y^2$
• D. $\frac{1}{2}ln|x^2+a^2|-ta{n}^{-1}\left(\frac{a}{x}\right)=c$ where $a=x+y^2$

Answer 1

The correct option is D $\frac{1}{2}ln|x^2+a^2|-ta{n}^{-1}\left(\frac{a}{x}\right)=c$ where $a=x+y^2$

Put $y^2=v-x.$

Differentiating, we get $y\frac{dy}{dx}=\frac{1}{2}(\frac{dv}{dx}-1)$

The given equation is

$y^2\left(y\frac{dy}{dx}\right)+x+y^2=0$

$\Rightarrow (v-x).\frac{1}{2}(\frac{dv}{dx}-1)+x+(v-x)=0$ (Putting $y^2=v-x$)

$\Rightarrow (v-x)\frac{dv}{dx}-(v-x)+2v=0$

$\Rightarrow \frac{dv}{dx}=\frac{v+x}{x-v}$ ...(1)

Thus the given equation is reduced to homogeneous form.

Now to solve it put $v=zx\Rightarrow \frac{dv}{dx}=z+x\frac{dz}{dx}$

$\therefore$ from (1), we get

$z+x\frac{dz}{dx}=\frac{z+1}{1-z}$

$\Rightarrow x\frac{dz}{dx}=\frac{z+1}{1-z}-z=\frac{1+z^2}{1-z}$

$\Rightarrow \frac{(1-z)dz}{1+z^2}=\frac{dx}{x}$

$\Rightarrow \frac{1}{1+z^2}dz-\frac{zdz}{1+z^2}=\frac{dx}{x}$

Now integrating, we get ${\tan }^{-1}z=\frac{1}{2}\log (1+z^2)=\log x+\log c$

where $c$ is an arbitrory constant

$\Rightarrow {\tan }^1\left(\frac{v}{x}\right)-\frac{1}{2}\log \left[\frac{(x^2+v^2)}{x^2}\right]=\log x+\log c[\because z=\frac{v}{x}]$

$\Rightarrow {\tan }^1\left(\frac{y^2+x}{x}\right)-\frac{1}{2}\log [x^2+{(y^2+x)}^2]$

$=\log c[\because v=y^2+x]$