DHS First-order Differential Equations (9649) (Revision Solutions)

1 H2 Double Math Differential Equations I (Solutions) Qn Suggested Solution 1(i) 22d2, 0 d yxy y x xx   22d2( ) ( ) ( )d vx vx x v vx xx  32 2 2 2 2d22 d vv x vx vx xx  32 2 2d2 d vvx v x xx   0,x thus 22 3 d( 1 ) d2 vx v xv x  Thus, 2d1 d2 vv xv x  (shown) (ii) 2d1 d2 vv xv x  2 2 ln 22 21 dd1 ln( 1) ln 1e xc v vxvx vx c v        2 11vA x   , A = ± e2c 2 11y Axx     22yx A x , where A is an arbitrary constant. (iii) When 2x , y = 0, 2A 22 2yx x 22 (1 )1yx (1, 0) O y x 1 Substitute y = vx dd dd yv x vxx 

2 Qn Su ggested Solution 2 dd s i ncos sin cos 1dd c o s yy tty t t ytt t     --- (*) I.F. = sin d ln coscosee c o s t t tt t    Multiply DE (*) throughout by I.F, dcos sin cosd yty t tt  d cos cosd yt tt  cos sinyt t c  GS: tan secyt c t , where c is an arbitrary constant. (i) 0, 0ty : 0c . Thus we sketch tanyt (ii) Sub ,6t  d 0d y t  into DE (*) : 1 13 3 yy   ,36ty   : 1232 33 c c    . Thus we sketch tan 2secyt t Question requested for sketch in a single diagram. t y O

3 Qn Suggested Solution 3(a) Method 1: Variable separable  d1c o s 1 d x x    d1c o s 1 d x x    1d 1 1d 1 c o s x x   2 11 d d1 2cos 2 xx  21ln 1 sec d 22x     ln 1 tan 2x C   tan 21e C x    tan 21 e , where e is an arbitrary constantCxA A     tan 21exc   , where c is an arbitrary constant. Method 2: Integrating factor  d1c o s 1 d x x   d1 d1 c o s 1 c o s xx   ---------- (*) 22 1 d1 1 d 2cos sec d tan1+cos 22 2 2Integrating factor = e e e e      Multiply (*) throughout by I, tan tan tan22 2d1ee e d1 c o s 1 c o s xx     tan tan22d1 eed1 c o sx       tan tan22 1ee d 1c o sx    tan tan22e e , where is an arbitrary constant xc c   tan 21e xc  

4 (b) 2d 2l nd yy yxxx 22 1d 1 2 lnd u xux u x u  d 2l nd uu xx x  ---------- (*) Integrating factor = 1 d e xx = lne x = 1 x Multiplying (*) throughout by I, 2 1d 2 lnd uu xxx x x  d1 2 lnd uxxx x   12l n du x xxx   2 ln2 2 xu Cx    2 lnux xC x   21 lnx xC xy    2 1 ln y Cx x x   , where C is an arbitrary constant Substitute 1y u 2 d1 d dd yu x ux 

5 Qn Su ggested Solution 4(a)(i) 2 d d yxy x  When 0y , d 0d y x  . Thus, 0y is a solution to the differential equation. When 0y , we have 2 1d 1 d y