(i) dydx=y tan x, y0=1(ii) 2xdydx=5y, y1=1(iii) dydx=2e2x y2, y0=-1(iv) cos ydydx=ex, y0=π2(v) dydx=2xy, y0=1(vi) dydx=1+x2+y2+x2y2, y0=1(vii) xydydx=x+2y+2, y1=-1(viii) dydx=1+x+y2+xy2 when y = 0, x = 0 [NCERT EXEMPLAR](ix) 2y+3-xydydx=0, y(1) = −2 [NCERT EXEMPLAR](x) extan y dx+2-exsec2y dy=0, y0=π4 [CBSE 2018]

(i) dydx=y tan x, y0=1(ii) 2xdydx=5y, y1=1(iii) dydx=2e2x y2, y0=-1(iv

1.	(i) dydx=y tan x, y0=1(ii) 2xdydx=5y, y1=1(iii) dydx=2e2x y2, y0=-1(iv) cos ydydx=ex, y0=π2(v) dydx=2xy, y0=1(vi) dydx=1+x2+y2+x2y2, y0=1(vii) xydydx=x+2y+2, y1=-1(viii) dydx=1+x+y2+xy2 when y = 0, x = 0 [NCERT EXEMPLAR](ix) 2y+3-xydydx=0, y(1) = −2 [NCERT EXEMPLAR](x) extan y dx+2-exsec2y dy=0, y0=π4 [CBSE 2018]
Answer» (i) $\frac{d y}{d x} = y \tan x, y (0) = 1$ (ii) $2 x \frac{d y}{d x} = 5 y, y (1) = 1$ (iii) $\frac{d y}{d x} = 2 e^{2 x} y^{2}, y (0) = - 1$ (iv) $\cos y \frac{d y}{d x} = e^{x}, y (0) = \frac{π}{2}$ (v) $\frac{d y}{d x} = 2 x y, y (0) = 1$ (vi) $\frac{d y}{d x} = 1 + x^{2} + y^{2} + x^{2} y^{2}, y (0) = 1$ (vii) $x y \frac{d y}{d x} = (x + 2) (y + 2), y (1) = - 1$ (viii) $\frac{d y}{d x} = 1 + x + y^{2} + x y^{2}$ when y = 0, x = 0 [NCERT EXEMPLAR] (ix) $2 (y + 3) - x y \frac{d y}{d x} = 0$ , y(1) = −2 [NCERT EXEMPLAR] (x) $e^{x} \tan y d x + (2 - e^{x}) \sec^{2} y d y = 0, y (0) = \frac{π}{4}$ [CBSE 2018]