The solution of $y'=1+x+y^2+xy^2,y(0)=0$ is:
Answer & explanation
Correct answer: option 4
$\int\frac{dy}{1+y^2}=\int(1+x)dx⇒tan^{-1}y=\frac{x^2}{2}+x+k;y=tan(\frac{x^2}{2}+x+k)$
y (0) = 0; k = 0; $y=tan(x+\frac{x^2}{2})$
The solution of $y'=1+x+y^2+xy^2,y(0)=0$ is:
Correct answer: option 4
$\int\frac{dy}{1+y^2}=\int(1+x)dx⇒tan^{-1}y=\frac{x^2}{2}+x+k;y=tan(\frac{x^2}{2}+x+k)$
y (0) = 0; k = 0; $y=tan(x+\frac{x^2}{2})$