Question

If $y$ is the soultion of the differential equation $y^3\frac{dy}{dx}+x^3=0,y\left(0\right)=1,$ the value of $y(-1)$ is

• A. $-2$
• B. $-1$
• C. $0$
• D. $1$

Answer 1

The correct option is C $0$

$y^3\frac{dy}{dx}+x^3=0\Rightarrow y^{3}dy=-x^{3}dx$ .... (i)
On integrating, $\frac{y^4}{4}=-\frac{x^4}{4}+C$ ...(ii)
Using $y\left(0\right)=1\Rightarrow C=\frac{1}{4}$
So, solution is $\frac{y^4}{4}=-\frac{x^4}{4}+\frac{1}{4}$
$\Rightarrow x^4+y^4=1$
Hence, $y=(1-x^4{)}^{1/4}$
$y(-1)=[1-(-1{)}^4{]}^{1/4}=0$