If dy/dx +(x/1+x^2)y = √x/√1+x^2; y(0)=0, then y(1) is

Question

If \(\frac{d y}{d x}+\left(\frac{x}{1+x^2}\right) y=\frac{\sqrt{x}}{\sqrt{1+x^2}} ; y(0)=0,\) then y(1) is

(1) \(\frac{2}{3}\)

(2) \(\frac{2}{\sqrt{3}}\)

(3) \(\frac{\sqrt{2}}{3}\)

(4) \(\sqrt{\frac{2}{3}}\)

Answer 1

Correct option is (3) \(\frac{\sqrt{2}}{3}\)    

\(\frac{d y}{d x}-\left(\frac{x}{1+x^2}\right) y=\frac{\sqrt{x}}{\sqrt{1+x^2}}\)   

\(\text { If }=e^{\int \frac{x}{1+x^2} d x}=e^{\frac{1}{2} \ln \left(1+x^2\right)}=\sqrt{1+x^2}\)   

Solution will be \(y \sqrt{1+x^2}=\int \frac{\sqrt{x}}{\sqrt{1+x^2}} \cdot \sqrt{1+x^2} d x \)