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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
For each of the folowing differential equations, find a particular solution satisfying the given condition: `(dy)/(dx) =-4xy^(2)," it being given that " y = 1 " when " x = 0.` |
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Answer» Correct Answer - `y=(1)/((2x^(2)+1))` `int (dy)/(y^(2))=int -4x dx rArr (-1)/(y) = -2x^(2)+C rArr y = (1)/((2x^(2)-C)) " " `...(i) Putting x = 0 and y = 1 in (i), we get C = -1. Hwence, `y = (1)/((2x^(2)+1)).` |
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| 2. |
Find the general solution of each of the following differential equations: `(dy)/(dx)=(1-cos x)/(1+cosx)` |
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Answer» Correct Answer - `y=2 "tan"(x)/(2)-x+C` `(dy)/(dx)=(1-cosx)/(1+cosx)=(2sin^(2)(x//2))/(2cos^(2)(x//2))="tan"^(2)(x)/(2)=("sec"^(2)(x)/(2)-1)` ` therefore int dy = int ("sec"^(2)(x)/(2)-1)dx.` |
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| 3. |
Find the general solution of each of the following differential equations: `(1)/(x)*(dy)/(dx)=tan^(-1)x` |
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Answer» Correct Answer - `y=(1)/(2)(x^(2)+1)tan^(-1)x-(1)/(2)x+C` `intdy = int (underset(I)(tan^(-1))x)underset(II)(x) dx rArr y=(tan^(-1)x)(x^(2))/(2)-int (1)/((1+x^(2)))*(x^(2))/(2)dx +C` `rArr y = (1)/(2)x^(2)(tan^(-1)x)-(1)/(2)int{((1+x^(2))-1)/((1+x^(2)))}dx +C.` `rArr y =(1)/(2)x^(2)(tan^(-1)x)-(1)/(2)int{1-(1)/((1+x^(2)))}dx +C`. |
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| 4. |
Find the general solution of each of the following differential equations: `ydx +(1+x^(2))tan^(-1)xdy =0` |
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Answer» Correct Answer - `ytan^(-1)x=C` `int (dx)/((1+x^(2))tan^(-1)x)+int(1)/(y)dy =log|C_(1)|.` Put `tan^(-1)x=t and (dx)/((1+x^(2)))=dt.` |
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| 5. |
Find the general solution of the differential equations `sec^2xtany dx+sec^2ytanx dy=0` |
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Answer» Correct Answer - `tan y = C (1-e^(x))` `int (sec^(2)x)/(tanx) dx +int (sec^(2)y)/(tany) dy = log |C_(1)|rArr log |tanx|+log |tan y| = log |C_(1)|` `therefore log |tan x tany| = log |C_(1)| rArr |tan x tany|=|C_(1)|.` `therefore tanx tan y = pm C_(1) = C.` |
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| 6. |
Find the general solution of each of the following differential equations: `(dy)/(dx) = sin^(3)x cos^(2)x+xe^(x)` |
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Answer» Correct Answer - `y=-(1)/(3) "cos"^(3)x+(1)/(5)cos^(5)x +xe^(x)-e^(x)+C` `int dy=int cos^(2)x(1-cos^(2)x)sinx dx +int xe^(x)dx+C` `rArr y = -int t^(2)(1-t^(2))dt+int xe^(x)dx+C, " where " t = cosx` `rArr y = -int t^(2)dt +int t^(4)dt +int underset(I)(x)underset(II)(e^(x))dx +C. ` |
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| 7. |
Find the general solution of each of the following differential equations: `(dy)/(dx)+sin(x+y)=sin(x-y)` |
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Answer» Correct Answer - `log|"cosec"y-coty|+2 sinx =C` `sin(x-y)-sin(x+y)=2cos{((x-y)+(x+y))/(2)}sin{((x-y)-(x+y))/(2)}` `=2cosx sin(-y)= -2 cosx siny.` `therefore int (1)/(siny)dy = -2 int cosx dx rArr int "cosec"y dy = -2sinx +C` `rArr log|"cosec"y-coty|= -2 sinx +C.` |
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| 8. |
Find the general solution of each of the following differential equations: `sin^(3)xdx -siny dy =0` |
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Answer» Correct Answer - `12 cosy = 9 cosx -cos3x +C` `siny dy = sin^(3)x dx = ((3sinx-sin3x))/(4)dx " " [because sin 3x = 3 sinx -4 sin^(3)x]` `therefore int siny dy =(3)/(4)intsinx dx -(1)/(4)int sin3x dx +C.` |
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| 9. |
Find the general solution of each of the following differential equations: `(1)/(x)cos^(2)y dy +(1)/(y)cos^(2)xdx=0` |
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Answer» Correct Answer - `2(x^(2)+y^(2))+2xsin2x+2y sin2y+cos2x+cos2y=C` `ycos^(2)y dy +xcos^(2)x dx =0 rArr y(2cos^(2)y)dy +x(2cos^(2)x)dx=0` `rArr y(1+cos2y)dy+x(1+cos2x)dx=0` `rArr int y dy +int y cos 2y dy +int x dx +int x cos2x dx = C` `rArr (x^(2))/(2)+(y^(2))/(2)+int underset(I)(y)underset(II)(cos 2y)dy+int underset(I)(x)underset(II)(cos 2x)dx =C.` |
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| 10. |
Find the general solution of each of the following differential equations: `(dy)/(dx)=(1+x^(2))(1+y^(2))` |
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Answer» Correct Answer - `tan^(-1)y=x+(x^(3))/(3)+C` `int(1)/((1+y^(2)))dy = int (1+x^(2))dx +C.` |
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| 11. |
Find the general solution of each of the following differential equations: `x^(4)(dy)/(dx)=-y^(4)` |
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Answer» Correct Answer - ` (1)/(x^(3))+(1)/(y^(3))= C` `(-1)/(y^(4))dy = (1)/(x^(4))dx rArr int x^(-4)dx +int y^(-4)dy =C_(1) rArr (1)/(x^(3))+(1)/(y^(3))= -3C_(1) = C.` |
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| 12. |
Find the general solution of each of the following differential equations: `(dy)/(dx) +((1+cos 2y))/((1-cos 2x)) =0` |
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Answer» Correct Answer - `tany =cot x +C` `(dy)/(dx)+(2cos^(2)y)/(2sin^(2)x)=0 rArr (dy)/(dx)+("cosec"^(2)x)/(sec^(2)y)=0 rArr int (sec^(2)y)dy +int "cosec"^(2)xdx =C.` |
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| 13. |
Find the general solution of each of the following differential equations: `(cosx)(dy)/(dx)+cos2x = cos 3x` |
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Answer» Correct Answer - `y=sin 2x -2 sinx -x +log |sec x +tan x |+C` `(dy)/(dx)=(cos3x-cos 2x)/(cosx)=((4cos^(3)x-3cosx)-(2cos^(2)x-1))/(cosx)` `rArr (dy)/(dx)=4cos^(2)x-2cosx-3+secx` ` rArr (dy)/(dx) =(4(1+cos2x))/(2)-2cosx-3 +secx =2 cos 2x -2cosx -1 +sec x.` |
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| 14. |
Find the general solution of each of the following differential equations: `x(x^(2)-x^(2)y^(2))dy +y(y^(2)+x^(2)y^(2))dx =0` |
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Answer» Correct Answer - `(-1)/(2y^(2))-(1)/(2x^(2))+log|(x)/(y)|=C` `x^(3)(1-y^(2))dy +y^(3)(1+x^(2))dx =0 rArr int ((1-y^(2)))/(y^(3))dy + int ((1+x^(2)))/(x^(3)) dx = C.` `therefore int (y^(-3)-(1)/(y)) dy +int (x^(-3)+(1)/(x))dx = C rArr (y^(-2))/(-2)-log|y|+(x^(-2))/(-2) +log|x|=C.` |
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| 15. |
Find the general solution of each of the following differential equations: `(dy)/(dx)=1+x+y+xy` |
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Answer» Correct Answer - `log|1+y|=x+(x^(2))/(2)+C` `(dy)/(dx) = (1+x)+y(1+x)=(1+x)(1+y) rArr int (dy)/((1+y))=int (1+x)dx +C.` |
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| 16. |
Find the general solution of each of the following differential equations: `(1-x^(2))dy +xy(1-y)dx =0` |
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Answer» Correct Answer - `y=C(1-y)sqrt(1-x^(2))` `int (1)/(y(1-y))dy + int (x)/((1-x^(2)))dx = log |C_(1)|` `rArr int {(1)/(y)+(1)/((1-y))}dy -(1)/(2)int (-2x)/((1-x^(2)))dx =log|C_(1)|` `rArr log|y|-log|1-y|-(1)/(2)log|1-x^(2)|=log|C_(1)| rArr log|(y)/((1-y)sqrt(1-x^(2)))|=log|C_(1)|` `rArr (y)/((1-y)sqrt(1-x^(2)))= pm C_(1)=C.` |
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| 17. |
Find the general solution of each of the following differential equations: `(x-1)(dy)/(dx)=2x^(3)y` |
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Answer» Correct Answer - `log|y| =(2x^(3))/(3) +x^(2)+2x+2log|x-1| +C` `(1)/(y)dy=(2x^(3))/((x-1))=2[x^(2)+x+1+(1)/((x-1))]["on dividing "x^(3) " by " (x-1)]` |
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| 18. |
Find the general solution of each of the following differential equations: `(1-x^(2))(1-y)dx = xy(1+y)dy` |
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Answer» Correct Answer - `log|x(1-y^(2))|=(x^(2))/(2)-(y^(2))/(2)-2y+C` `int ((1-x^(2)))/(x)dx = int(y(1+y))/((1-y))dy rArr int ((1)/(x)-x)dx=int((y^(2)+y))/((-y+1))dy` `therefore int ((1)/(x)-x)dx=int(-y-2+(2)/(1-y))dy " "["on dividing "(y^(2)+y)"by"(-y+1)].` |
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| 19. |
Find the general solution of each of the following differential equations: `(dy)/(dx)=e^(x+y)` |
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Answer» Correct Answer - `e^(x)+e^(-y) =C` `(dy)/(dx)=e^(x)*e^(y) rArr (1)/(e^(y)) dy = e^(x)dx rArr int e^(-y)dy = int e^(x)dx +C` |
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| 20. |
Find the general solution of each of the following differential equations: `(dy)/(dx)=(3e^(2x)+3e^(4x))/(e^(x)+e^(-x))` |
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Answer» Correct Answer - `y=e^(3x)+C` `(dy)/(dx)=(3e^(2x)(1+e^(2x)))/((e^(x)+(1)/(e^(x))))=(3e^(3x)(1+e^(2x)))/((1+e^(2x)))=3e^(3x).` |
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| 21. |
Find the general solution of each of the following differential equations: `(dy)/(dx)=e^(x+y)+e^(x-y)` |
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Answer» Correct Answer - `tan^(-1)(e^(y))=e^(x)+C` `int(e^(y))/((e^(2y)+1))dy =int e^(x)dx rArr int (dt)/((t^(2)+1))=e^(x)+C." where " e^(y)=t` `rArr tan^(-1)(t)=e^(x)+C rArr tan^(-1)(e^(y))=e^(x)+C.` |
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| 22. |
Find the general solution of each of the following differential equations: `3e^(x)tany dx +(1-e^(x))sec^(2)y dy=0` |
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Answer» Correct Answer - `tany = C (1-e^(-x))^(3)` `int (3e^(x))/((1-e^(x)))dx+int(sec^(2)y)/(tany)dy=log|C_(1)|` `rArr -3int (-e^(x))/((1-e^(x)))dx+int (sec^(2)y)/(tany)dy =log|C_(1)|rArr -3log|1-e^(-x)|+log|tan y|=log|C_(1)|` `rArr log|(1-e^(-x))^(-3)tany|=log|C_(1)|rArr (tan y)/((1-e^(-x))^(3)) = pm C_(1)=C.` |
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| 23. |
Find the general solution of each of the following differential equations: `(e^(x)+e^(-x))dy-(e^(x)-e^(-x))dx=0` |
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Answer» Correct Answer - `y = log (e^(x)+e^(-x)) +C` `dy-((e^(x)-e^(-x)))/((e^(x)+e^(-x)))dx=0 rArr int dy - int (e^(x)-e^(-x))/(e^(x)+e^(-x))dx + C rArr y - log(e^(x)+e^(-x)) =C.` |
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| 24. |
Find the general solution of each of the following differential equations: `(dy)/(dx)=e^(x-y)+x^(2)e^(-y)` |
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Answer» Correct Answer - `e^(y)=e^(x)+(x^(3))/(3)+C` `(dy)/(dx)=e^(x)*e^(-y)+x^(2)e^(-y)=(e^(x)+x^(2))e^(-y) rArr int e^(y) dy = int (e^(x)+x^(2))dx + C. ` |
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| 25. |
Find the general solution of each of the following differential equations: `e^(x) tan y dx +(1-e^(x))sec^(2)y dy =0` |
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Answer» Correct Answer - `tany=C(1-e^(x))` `-int (-e^(x))/((1-e^(x)))dx + int (sec^(2)y)/(tan y) dy = log |C_(1)| rArr logh|tan y|-log |1-e^(x)|=log |C_(1)|.` ` therefore log |(tany)/((1-e^(x)))|=log |C_(1)| rArr (tan y)/((1-e^(x))) = pm C_(1) = C.` |
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| 26. |
Findthe particular solution of the differential equation `log(dy)/(dx)=3x+4y`given that `y" "=" "0`when`x" "=" "0`. |
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Answer» Correct Answer - `4e^(3x)+3e^(-4y)=7` `(dy)/(dx)=e^(3x+4y)=e^(3x)*e^(4y) rArr int e^(3x)dx = int e^(-4y)dy` `therefore (e^(3x))/(3)=(e^(-4y))/(-4)+C. " " `...(i) Putting x = 0 and y = 0 in (i), we get `C=((1)/(3)+(1)/(4))=(7)/(12).` `therefore (e^(3x))/(3)=(e^(-4y))/(-4)+(7)/(12) rArr 4e^(3x)+3e^(-4y)=7.` |
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| 27. |
Find the general solution of the differential equation `(dy)/(dx) = log(x+1).` |
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Answer» We have , `(dy)/(dx)=log(x+1)` `rArr dy=log(x+1)dx` `rArr int dy = int log(x+1)dx` [integrating both sides] `rArr y=int {log(x+1)*1}dx + C, ` where C is an arbitrary constant `={log(x+1)*x}-int(1)/((x+1)) *x dx + C " " `[integrating by parts] `=x log (x+1) - int ((x+1)-1)/((x+1)) dx + C` `=x log (x+1) - int {1-(1)/((x+1)) } dx + C` ` = x log (x+1)-x + log(x+1) + C` `=(x+1)log(x+1)-x+C.` Hence, `y=(x+1)log(x+1)-x+C` is the required solution. |
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| 28. |
Find the general solution of the differential equation `(1+x^(2))(dy)/(dx)-x=2tan^(-1)x.` |
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Answer» The given differential equation may be written as `(dy)/(dx)=(2tan^(-1)x)/((1+x^(2)))+(x)/((1+x^(2)))` `rArr dy={(2tan^(-1)x)/((1+x^(2)))+(x)/((1+x^(2)))}dx " " ` [separating the variables] `rArr int dy=int (2tan^(-1)x)/((1+x^(2)))dx+int (x)/((1+x^(2)))dx + C, " " ` where C is an arbitrary constant `rArr y= 2int t dt + (1)/(2) int (2x)/((1+x^(2))) dx +C ` [putting `tan^(-1)x=t and (1)/((1+x^(2))) dx = dt ` in 1st integral] `rArr y=t^(2)+(1)/(2)log|1+x^(2)| + C` `rArr y=(tan^(-1)x)^(2)+(1)/(2)log|1+x^(2)|+C.` Hence, `y=(tan^(-1)x)^(2)+(1)/(2)log|1+x^(2)|+C` is the required solution. |
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| 29. |
Find the general solution of each of the following differential equations: `(dy)/(dx)+sqrt((1-y^(2))/(1-x^(2)))=0` |
| Answer» Correct Answer - `sin^(-1)y + sin^(-1)x=C` | |
| 30. |
The volume of sphericalballoon being inflated changes at a constant rate. If initially its radius is3 units and after 3 seconds it is 6 units. Find the radius of balloon after tseconds. |
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Answer» Correct Answer - `r=(63t+27)^(1//3)` The volume of a spherical balloon of radius r is given by `V=(4)/(3) pi r^(3).` Now, `(dV)/(dt) = -k, " where " k gt 0 " " ` [note that V is decreasing] `rArr (d)/(dt)((4)/(3)pi r^(3))= -k rArr (4pr^(2))(dr)/(dt) = -k` `rArr int (4pir^(2))dr=int (-k)dt` `rArr (4)/(3) pi r^(3) = -kt +C " " `...(i), where C is an arbitrary constant. Putting t = 0 and r = 3 in (i), we get C = 36 `pi`. `therefore (4)/(3)pir^(3)= -alpha t +36 pi. " " `...(ii) It is being given that when t = 3, then r = 6. Putting t = 3 and r = 6 in (ii), we get k = -84`pi`. Putting k = -84`pi`. in (ii) , we get `r^(3)=(63t +27) rArr r = (63t +27)^(1//3).` |
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| 31. |
Solve the differential equation `(y)/(dx)=y sin 2x, " given that " y(0) =1.` |
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Answer» Correct Answer - `y=e^(sin^(2)x)` `int (1)/(y)dy = int sin2x dx +C rArr log |y| = -(1)/(2)cos2x +C." " `...(i) Putting x = 0 and y = 1 in (i), we get C = `(1)/(2).` `therefore log |y|=(1)/(2)(1-cos2x)=((1)/(2) xx 2 sin^(2)x)=sin^(2)x.` Hence, `y = e^(sin^(2)x).` |
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| 32. |
Solve `(x^(3)+x^(2)+x+1)(dy)/(dx) =2x^(2)+x, " given that " y=1 " when " x =0.` |
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Answer» Correct Answer - `y=(1)/(2){log|x+1|+(3)/(2)log (x^(2)+1)-tan^(-1)x}+1` `dy=(1)/(2){(1)/((x+1))+(3x-1)/((x^(2)+1))}dx " " `[by partial fractions] `rArr int dy=(1)/(2) int {(1)/((x+1))+(3)/(2)*(2x)/((x^(2)+1))-(1)/((x^(2)+1))}dx +C` `rArr y =(1)/(2) {log |x+1|+(3)/(2)log |x^(2)+1|-tan^(-1)x}+C.` When x = 0 and y = 1, then C = 1. |
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| 33. |
Solve the differential equation `(x+1)(dy)/(dx) =2xy, " given that " y(2)=3.` |
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Answer» Correct Answer - `y(x+1)^(2)=27e^((2x-4))` `int (1)/(y)dy =int (2x)/((x+1))dx = 2int ((x+1)-1)/((x+1))dx =2int {1-(1)/((x+1))}dx` ` rArr log |y|=2x - 2 log |x+1|+log |C_(1)|` `rArr log |y|=log |e^(2x)|-log|(x+1)^(2)|+log |C_(1)|` `rArr log|(y(x+1)^(2))/(e^(2x))|=log|C_(1)|rArr (y(x+1)^(2))/(e^(2x))=pm C_(1)=C` (say). Then, `y(x+1)^(2)=Ce^(2x).` Putting x =2 and y =3 in (i), we get `C=27e^(-4)`. `therefore y(x+1)^(2)=27e^(2x-4)` is the required solution. |
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| 34. |
Solve `(dy)/(dx) =y cot 2x, " given that " y =2 " when " x =(pi)/(4)`. |
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Answer» Correct Answer - `y=2sqrt(sin 2x)` `int (dy)/(dx)=int (cos 2x)/(sin2x)dx rArr log |y| = (1)/(2) log |sin 2x|+log |C_(1)|` `therefore (y)/(sqrt(sin2x))=C_(1) rArr (y)/(sqrt(sin2x))=pm C_(1)=C "(say). " ` ...(i) Putting `x = (pi)/(4)` and y = 2 in (i), we get C = 2. Hence, `y=2sqrt(sin 2x).` |
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| 35. |
Find the equation of the curve that passes through the point (1, 2) and satisfies the differential equation `(dy)/(dx) =(-2xy)/((x^(2)+1)).` |
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Answer» We have `(dy)/(dx) =(-2xy)/((x^(2)+1))` `rArr (dy)/(y)=(-2x)/((x^(2)+1))dx" " `[on separating the variables] `rArr int (dy)/(y)=int(-2x)/((x^(2)+1))dx" " `[integrating both sides] `rArr log y = -log(x^(2)+1)+logC,` where log C is an arbitrary constant `rArr log y + log (x^(2)+1)=log C` `rArr log {y(x^(2)+1)}=log C` `rArr y(x^(2)+1)=C " " `...(i) Now, it is given that the curve passes through (1, 2). So, putting x = 1 and y = 2 in (i), we get C = 4 `therefore y=(x^(2) +1)=4` is the required equation of the curve. |
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| 36. |
Find the equation of a curve which passes through the origin and whose differential equation is `(dy)/(dx) =e^(x) sin x`. |
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Answer» Correct Answer - `2y=e^(x)(sinx-cosx)+1` Use the formula `inte^(ax)sin bx dx = (e^(ax)(asinx bx-b cosbx))/((a^(2)+b^(2))).` Put `a = 1 and b = 1`. |
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| 37. |
Find the equation of a curve, passes through `(-2,3)` at which the slope of tangent at any point `(x,y)` is `(2x)/(y^(2))`. |
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Answer» We know the slope of a curve at a point (x, y) is `(dy)/(dx)`. `therefore (dy)/(dx)=(2x)/(y^(2)) " " ` …(i) `rArr y^(2)dy = 2x dx " " `[separating the variables] `rArr int y^(2)dy = int 2x dx ` `rArr (1)/(3)y^(3) = x^(2) +C " " `...(ii) where C is a constant. Thus, (ii) is the equation of the curve whose differential equation is given by (i) . Since the given curve passes through the point (-2, 3), we have `C=((1)/(3)xx 27)-(-2)^(2)=(9-4)=5.` Hence , the required equation of the curve is ` (1)/(3) y^(3) = x^(2)+5 rArr y^(3) =3x^(2)+15`. |
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| 38. |
A curve passes through the point (-2, 1) and at any point (x, y) of the curve, the slope of the tangent is twice the slope of the line segment joining the point of contact to the point (-4, -3). Find the equation of the curve. |
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Answer» Correct Answer - `y+3=(x+4)^(2)` `(dy)/(dx) = (2(y+3))/((x+4))rArr int (1)/((y+3))dy =2 int (1)/((x+4))dx` `therefore log |y+3|=2log|x+4|+log|C_(1)| rArr ((y+3))/((x+4)^(2)) =pm C_(1)=C.` Putting x = -2 and y = 1, we get C = 1. |
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| 39. |
Find the equation of the curve passing through thepoint (0, -2) given that at any point `(x , y)`on the curve the product of the slope of its tangentand `y`coordinate of the point is equal to thex-coordinate of the point. |
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Answer» Correct Answer - `y^(2)=x^(2)+4` `y(dy)/(dx) =x rArr inty dy = int x dx rArr (y^(2))/(2) = (x^(2))/(2)+C.` Put x = 0 and y = 2 to get C = 2. |
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| 40. |
Find the general solution of each of the following differential equations: `e^(2x-3y)dx+e^(2y-3x)dy =0` |
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Answer» Correct Answer - `e^(5x)+e^(5y)=C` `e^(2x)*e^(-3y) dx +e^(2y)*e^(-3x)dy = 0 rArr int e^(5x)dx + int e^(5y)dy = C.` |
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| 41. |
Find the general solution of each of the following differential equations: `(dy)/(dx)+(cosxsiny)/(cosy)=0` |
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Answer» Correct Answer - `log |sin y| +sin x=C` `int cot y dy +int cosx dx =C.` |
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| 42. |
Find the general solution of each of the following differential equations: `cosx(1+cosy)dx -siny (1+sinx)dy =0` |
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Answer» Correct Answer - `(1+sin x)(1+cosy) =C` `int (cosx)/((1+cosx))dx + int (-siny)/((1+cosy))dy = log |C_(1)|` `rArr log |1+sinx|+log|1+cosy|=log |C_(1)|rArr log |(1+sinx)(1+cosy)=log |C_(1)|` `rArr |(1+sinx) (1+cosy)|=C_(1) rArr (1+sinx)(1+cosy) = pm C_(1) = C.` |
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