Suppose y is a function of x that satisfies dy/dx = √(1-y²)/x² and y=0 at x = 2/π, then y(3/π) is equal to:

  $\frac{dy}{dx}-\left(\frac{sin2x}{1+co{s}^{2}x}\right)y=\frac{sinx}{1+co{s}^{2}x}$

IF = $e^{-{\displaystyle \int \frac{sin2xdx}{1+co{s}^{2}x}}}$  

$=e^{ln\left(1+co{s}^{2}x\right)}=\left(1+co{s}^{2}x\right)$        

So, y(1 + cos² x) = ${\displaystyle \int \frac{sinx}{\left(1+co{s}^{2}x\right)}\cdot \left(1+co{s}^{2}x\right)dx}$  

y(1 + cos² x) = – cos x + c

$?$      y(0) = 0

0 = – 1 + c

-> c = 1

$y=\frac{1-cosx}{1+co{s}^{2}x}$   

Now, $y\left(\frac{\pi }{2}\right)=1$  

(t + 1)dx = (2x + (t + 1)³)dt

$\frac{dx}{dt}-\frac{2x}{t+1}={\left(t+1\right)}^2$

I.F. $=e^{{\displaystyle \int -\frac{2}{t+1}dt}}=\frac{1}{{\left(t+1\right)}^2}$  

Solution is

$\frac{x}{{\left(t+1\right)}^2}={\displaystyle \int 1dt}$  

x = (t + c) (t + 1)²

$?$ x (0) = 2 then c = 2

x = (t + 2) (t + 1)²

 x (1) = 12

$\left(4+x^2\right)dy-2x\left(x^2+3y+4\right)dx=0$

$\Rightarrow \frac{dy}{dx}=\left(\frac{6x}{x^2+4}\right)y+2x$

$e^{-3\mathcal{l}n\left(x^2+4\right)}=\frac{1}{{\left(x^2+4\right)}^3}$

so $\frac{y}{{\left(x^2+4\right)}^3}={\displaystyle \int \frac{2x}{{\left(x^2+4\right)}^3}dx+c}$

$\Rightarrow y=-\frac{1}{2}\left(x^2+4\right)+c{\left(x^2+4\right)}^3$

When x = 0, y = 0 gives $c=\frac{1}{32}$

So, for x = 2, y = 12

$\frac{dy}{dx}+2ytanx=sinx,\therefore I.F.e^{{\displaystyle \int 2tanxdx}}=se{c}^{2}x$  

= cos x – 2 cos² x = $-2{\left(cosx-\frac{1}{4}\right)}^2+\frac{1}{8}\therefore y_{max}=\frac{1}{8}$

. xdy - ydx - x² (xdy + ydx) + 3x? dx = 0
⇒ (xdy - ydx)/x² - (xdy + ydx) + 3x²dx = 0 ⇒ d (y/x) - d (xy) + d (x³) = 0
Integrate both side, we get
y/x - xy + x³ = c
Put x = 3, y = 3
⇒ 1 - 9 + 27 = c
c = 19
Put x = 4
y/4 - 4y = 19 - 64
⇒ y = 12