The solution of $y'=1+x+y^2+xy^2,y(0)=0$

CUET Preparation Today

Start Free Mocks

Home Questions X7K3Q5ASJN

Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Differential Equations

Question:

The solution of $y'=1+x+y^2+xy^2,y(0)=0$ is:

Options:

$y^2=exp.(x+\frac{x^2}{2})-1$

$y^2=1+c\,exp.(x+\frac{x^2}{2})$

$y=tan(c+x+x^2)$

$y=tan(x+\frac{x^2}{2})$

Correct Answer:

$y=tan(x+\frac{x^2}{2})$

Explanation:

$\int\frac{dy}{1+y^2}=\int(1+x)dx⇒tan^{-1}y=\frac{x^2}{2}+x+k;y=tan(\frac{x^2}{2}+x+k)$

y (0) = 0; k = 0; $y=tan(x+\frac{x^2}{2})$