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Question

(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

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Solution

(i) dydx=y tanx, y0=11ydy =tan x dxIntegrating both sides, we get1ydy =tan x dxlog y=log sec x+C .....(1)We know that at x=0 and y=1.Substituting the values of x and y in (1), we getlog 1 =log 1 +CC=0Substituting the value of C in (1), we get

log y=log sec x+0y= sec xHence, y=sec x, where x-π2,π2, is the required solution.


(ii) 2xdydx=5y, y1=12ydy=5x dxIntegrating both sides, we get 21ydy =51x dx2log y=5log x+ C .....(1)We know that at x=1 and y=1.Substituting the values of x and y in (1), we get2log 1 =5log 1 +CC=0Substituting the value of C in (1), we get

2 log y=5 log x+0y=x52Hence, y=x52 is the required solution.

(iii) dydx=2e2xy2, y0=-11y2dy=2e2x dxIntegrating both sides, we get1y2dy =2e2x dx-1y=e2x+ C .....(1)We know that at x=0, y=-1.Substituting the values of x and y in (1), we get1=1+CC=0Substituting the value of C in (1), we get

-1y=e2xy=-e-2xHence, y=-e-2x is the required solution.

(iv) cos ydydx=ex, y0=π2cos y dy=ex dxIntegrating both sides, we getcos y dy=ex dxsin y=ex+C .....(1)We know that at x=0, y=π2.Substituting the values of x and y in (1), we get1=1+CC=0Substituting the value of C in (1), we get

sin y=exy=sin-1exHence, y=sin-1ex is the required solution.

(v) dydx=2xy, y0=11ydy=2x dxIntegrating both sides, we get 1ydy=2x dxlog y=x2+C .....(1)We know that at x=0, y=1.Substituting the values of x and y in (1), we get0=0+CC=0Substituting the value of C in (1), we get log y=x2y=ex2Hence, y=ex2 is the required solution.

(vi) dydx=1+x2+y2+x2y2, y0=1dydx=1+x21+y2dy1+y2=1+x2 dxIntegrating both sides, we get dy1+y2=1+x2 dxtan-1y=x+x33+C .....(1)We know that at x=0, y=1.Substituting the values of x and y in (1), we getπ4=0+0+CC=π4Substituting the value of C in (1), we get tan-1y=x+x33+π4Hence,tan-1y=x+x33+π4 is the required solution.

(vii) xydydx=x+2y+2, y1=-1yy+2dy=x+2xdxy+2-2y+2dy=x+2xdx1-2y+2dy=1+2xdxIntegrating both sides, we get 1-2y+2dy=1+2xdxy-2log y+2=x+2log x+C .....(1)We know that at x=1, y=-1.Substituting the values of x and y in (1), we get-1-2log 1=1+2log 1+C-1=1+CC=-2Substituting the value of C in (1), we get y-2log y+2=x+2log x-2Hence, y-2log y+2=x+2log x-2 is the required solution.

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