InterviewSolution
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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. |
`x(dy)/(dx)-y=x+1` |
| Answer» Correct Answer - ` y = x log x - 1 + Cx ` | |
| 2. |
`2x(dy)/(dx)+y=6x^3` |
| Answer» Correct Answer - ` y = (6)/(7) x ^(3) + (C)/(sqrt x)` | |
| 3. |
` x (dy)/(dx) - y = 2x ^(3)` |
| Answer» Correct Answer - `y = x ^(3)+ Cx ` | |
| 4. |
` (dy)/(dx) = y tan x - 2 sinx ` |
| Answer» Correct Answer - ` 2y cos x = cos 2x + C ` | |
| 5. |
` (sinx ) (dy)/(dx) + (cos x )y = cos x sin ^(2) x ` |
| Answer» Correct Answer - ` y sin x = (1)/(3) sin ^(3) x + C ` | |
| 6. |
The solution of the differential equation `(dy)/(dx)+2ycotx=3x^2cosec^2x` is |
| Answer» Correct Answer - ` y sin ^(2) x = x ^(3) +C ` | |
| 7. |
Solve ` x (dy)/(dx) + 2y = x cosx ` |
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Answer» The given differential eqution may be written as ` (dy)/(dx) + (2)/(x) * y = cos x " "` ... (i) This is of the form ` (dy)/(dx) +Py =Q`, where `P = (2)/(x) and Q = cos x ` Thus, the given equation is linear. `IF = e ^(int P dx ) = e ^( int (2)/(x) dx) = e ^(2 log x ) = e ^( log (x ^(2)) ) = x ^(2)` So, the required solution is `y xx IF = int {Qxx (IF)}dx + C`, i.e., `yx ^(2) = int x ^(2) cos x dx + C ` ` = x ^(2) sinx - int 2x sin x dx + C ` [integrating by parts] ` " " = x ^(2) sinx - 2 * [ x(-cos x ) - int 1* (-cos x ) dx ] + C ` [ integrating by parts] ` " " = x ^(2) sin x + 2x cos x - 2 sinx +C ` Hence, `y = sin x + (2)/(x) cos x - (2)/(x ^(2)) sinx + (C)/( x ^(2)) ` is the required solution. |
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| 8. |
Solve: `dy / dx + y secx = tanx` |
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Answer» The given equation is of the form ` (dy)/(dx) + Py = Q`, where ` P = sec x and Q = tan x ` Thus, the given equation is linear. `IF =e ^( int Pdx) = e ^( int sec x dx ) = e ^(log (sec x + tan x ) ) = (sec x + tanx )` So, the required solution is ` y xx IF = int { Q xx( IF ) }dx + C`, i.e, `y (secx + tan x ) = int tan x (sec x + tan x ) dx + C ` ` " " = int sec x tan x dx = int tan ^(2) x dx + C ` ` " " = sec x + int (sec ^(2) x - 1 ) dx + C ` ` " " =sec x + tanx - x + C`. Hence, ` y (sec x +tan x ) =sec x + tanx - x+ C ` is the required solution. |
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| 9. |
Find the particularsolution of the differential equation.`(dy)/(dx)+ycotx=4x cos e c x , (x!=0), `given that `y=0`when `x=pi/2dot` |
| Answer» Correct Answer - `y sin x = 2x ^(2) - (1)/(2) pi^(2)` | |
| 10. |
Solve the following differential equations:`(dy)/(dx)+(4x)/(x^2+1)y+1/((x^2+1)^2)=0` |
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Answer» Correct Answer - `y= (-x)/((x^(2) + 1 ) ^(2)) + (C)/(( x^(2) + 1 ) ^(2))` |
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| 11. |
`(1 - x ^(2)) (dy)/(dx) + xy = x sqrt(1-x^(2))` |
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Answer» Correct Answer - `y = (-1)/(2) sqrt (1 - x ^(2)) log ( 1- x ^(2)) + C sqrt ( 1- x ^(2))` ` (dy)/(dx) + (x)/( (1- x ^(2))) * y = (x)/(sqrt (1 - x ^(2)))` ` IF = e ^(-(1)/(2)int (-2x)/((1- x ^(2))) dx) = e^(- (1)/(2) log ( 1- x ^(2)) = e ^(log"" (1)/(sqrt (1- x ^(2)))) = (1)/(sqrt (1- x ^(2))` `therefore ` its solution is `y xx (1)/(sqrt (1- x ^(2))) = int (x)/(sqrt (1 - x ^(2))) * (1)/(sqrt (1- x ^(2))) dx = (-1)/(2) int (-2x )/(( 1 - x ^(2))) dx + C ` `" " = - (1)/(2) log (1 - x ^(2)) + C`. |
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| 12. |
` (dy)/(dx) + y cot x = sin 2x ` |
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Answer» Correct Answer - ` y sinx = (2)/(3) sin ^(3) x +C ` `IF = e ^(int cot x dx ) = e ^( log sinx ) = sinx ` `therefore y xx sinx =int sin x sin 2x dx +C ` `" " = 2 int sin ^(2) x cos x dx +C = 2 int t ^(2) dt +C, ` where ` sin x = t ` |
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| 13. |
Find the particular solution of the differential equation ` x(dy)/(dx) - y = (x + 1 ) e^(-x) ` given that `y = 0 ` when ` x = 1 `. |
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Answer» The given differential equation may be written as ` (dy)/(dx) - (1)/(x) * y = ((x + 1) e ^(-x))/( x )`...(i) This is of the form ` (dy)/(dx) + Py = Q`, where `P = (-1)/( x ) and Q = ((x + 1 ) e ^(-x))/( x )` Thus, the given differential equation is linear. `IF = e ^(int Pdx) = e ^(-int (1)/(x) dx)= e ^(log (1//x)) = (1)/(x)` So, its solution is given by `y xx IF = int (Q xx IF ) dx + C`, i.e., `y xx (1)/(x) = int ((x + 1 ) ^(-x))/( x ^(2)) dx + C `. `rArr y xx (1)/(x) = int underset ("I")( (1)/(x)) underset("II") e ^(-x) dx = int (1)/(x ^(2)) e ^(-x) dx + C ` `rArr y xx (1)/(x) = (1)/(x) * (- e ^(-x)) + int (1)/(x ^(2) (-e^(-x)) dx) + int (1)/(x ^(2)) e ^(-x) dx + C` `rArr (y)/(x)= (-e^(-x))/( x )+C` `rArr y = - e ^(-x) +Cx" " `...(iii) It is being given that when ` x = 1, `then `y =0 `. Putting `x = 1 and y = 0` in (ii) , we get `C = e ^(-1)`. `therefore y = - e ^(-x) + xe^(-1) ` is the required solution. |
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| 14. |
`(1-x^2)(dy)/(dx)+x y=a x` |
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Answer» Correct Answer - `y = a + c sqrt (1 - x ^(2))` ` (dy)/(dx) + (x)/(( 1- x ^(2))) *y = (ax)/((1-x ^(2)))` `IF = e ^(-(1)/(2) int (-2x)/((1 - x ^(2))) dx )= e ^(- (1)/(2) log (1 - x ^(2)) ) = (1)/(sqrt (1-x ^(2))` `therfore` Its solution is given by ` yxx(1)/(sqrt (1- x ^(2))) = int (1)/(sqrt (1- x ^(2))) xx (ax)/((1- x ^(2)))dx + C ` `" " = - (a)/(2) int (-2x )/((1 - x ^(2))^(3//2)) dx +C = - (a)/(2) int (dt)/(t^(3//2)) + C`, where ` (1- x ^(2)) = t ` `" " = - (a)/(2) int t^(-3//2) dt + C = (a)/(sqrtt) +C = (a)/(sqrt (1 - x ^(2))) +C ` |
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| 15. |
` (dy)/(dx) + 2xy = x`, given that `y =1` when ` x =0` |
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Answer» Correct Answer - ` 2y = 1 +e ^(-x^(2)) ` `IF = e ^(int 2 x dx ) = e ^(x^(2)) ` `therefore y xx e ^( x ^(2)) = int x e ^( x ^(2)) dx = (1)/(2) int e ^(t) dt +C`, where ` x ^(2) = t ` `" " = (1)/(2) e ^(x ^(2)) +C`. ` therefore y = (1)/(2) + ( C)/( e ^( x ^(2))) ` Putting `y = 1 and x = 0`, we get `C = (1)/(2)`. |
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| 16. |
`( 1+x ^(2)) dy + 2xy dx =cot x ` |
| Answer» Correct Answer - ` y ( 1+ x ^(2)) = log |sin x | + C` | |
| 17. |
The general solution of the DE ` (dy)/(dx) = ( y^(2)- x ^(2))/( 2xy )` isA. ` x ^(2) - y ^(2) = C _1 x `B. `x ^(2) + y ^(2) = C_1 y `C. `x ^(2) + y ^(2) = C_1 x `D. none of these |
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Answer» Correct Answer - C The given De is homogeneous. Put `y = v x and (dy)/(dx) = v + x (dv)/(dx) `. ` therefore int ( 2v)/( ( 1+ v ^(2))) dv = - int (1)/(x) dx rArr log ( 1 + v ^(2) ) + log x = log C ` ` therefore x ^(2) + y^(2) = C_1 x`. |
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| 18. |
Solve `(x^2+1)dy/dx+2xy=sqrt(x^2+4)` |
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Answer» The given differential equation may be written as ` (dy)/(dx) + (( 2x)/( x ^(2) + 1 )) y = ( sqrt ( x ^(2) + 4))/( (x ^(2) + 1)) " " ` ... (i) This is of the form ` (dy)/(dx) + P y = Q,` `" " ` where ` P= (2x)/(( x ^(2) + 1 )) and Q = (sqrt (x ^(2) + 4))/( ( x ^(2) + 1 ))` Thus, the given differential equation is linear. `IF = e ^( int Pdx)= e ^( int (2x)/((x ^(2) + 1 )) dx) = e ^( log (x ^(2) + 1 )) = (x ^(2)+ 1)` So, the required solution is given by `y xx IF = int { Q xx IF} dx + C`, i.e., `y (x^(2) + 1 ) = int(sqrt (x^(2) + 4))/( (x^(2) + 1 )) xx (x ^(2) + 1) dx` `rArr y (x ^(2) + 1 ) = int sqrt ( x^(2) + 4) dx ` ` " "= (1)/(2) x sqrt (x^(2) + 4) + (1)/(2) xx 2^(2) xx log |x + sqrt (x ^(2) + 4)| + C` `" " = (1)/(2) x sqrt (x^(2) + 4) + 2log |x + sqrt (x ^(2) + 4)| + C`. Hence, `y (x ^(2) + 1 ) = (1)/(2) x sqrt ( x ^(2) + 4) + 2log | x + sqrt (x ^(2) + 4) | + C ` is the required solution. |
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| 19. |
`(1+x^2)dy/dx-2xy=(x^2+2)(x^2+1)` |
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Answer» Correct Answer - `y = (x ^(2) + 1 ) ( x + tan ^(-1) x + C) ` Write `(x ^(2) + 2) = { (x ^(2) + 1 ) + 1 }` |
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| 20. |
Find the particular solution of the differential equation ` cos x (dy )/(dx) + y = sin x `, given that ` y = 2 `when ` x = 0 `. |
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Answer» The given differential equation may be written as ` (dy )/(dx) + (sec x )y = tan x" "`... (i) This is of the form ` (dy)/(dx) + Py = Q`, where `P = secx and Q = tanx ` Thus, the given differential equation is linear. ` IF = e ^(int Pdx) = e ^(int sec x dx ) = e ^(log (sec x + tan x )) = (sec x + tanx )` So, its solution is given by `y xx IF = int (Q xx IF) dx + C ` i.e. `y (sec x + tan x ) = int (tanx (secx + tanx )dx + C ` ` " " = int sec x tanx dx + int tan^(2) x dx + C ` `" " = sec x + int (sec ^(2) x - 1 ) dx + C ` ` " " = sec x + int sec ^(2) x dx - int dx +C ` ` " " = sec x + tanx - x + C`. Thus, `y (sec x + tanx ) = sec x + tan x - x + C " " ` ... (ii) It is given that what ` x = 0 `, then `y = 2 ` ` therefore ` putting ` x = 0 and y = 2 ` in (ii), we get ` C = 1 ` Hence, `y (sec x + tanx ) = secx + tanx - x + 1` is the required solution. |
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| 21. |
Solve the differential equation ` (cos ^(2) x ) (dy)/(dx) + y = tan x ( 0 le x lt (pi)/(2))` |
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Answer» The given differential equation may be written as ` (dy)/(dx) + (sec ^(2) x )y = (sec ^(2) x ) tan x " " ` ... (i) This is of th form ` (dy)/(dx) + Py = Q`, where `P= sec ^(2) x and Q = sec ^(2) x tanx `. Thus, the given differential equation is linear. `IF = e ^(int Pdx) = e ^(int sec ^(2)x dx ) = e ^(tanx )` So, its solution is given by `y xx IF = int ( Q xx IF )dx + C ` i.e, `yxx e ^(tanx ) = int e ^(tanx ) (sec ^(2) x ) tan x dx + C ` `" " = int underset("I")t underset ("II") (e^(t)) dt, ` where ` tanx = t and sec ^(2) x d x = dt ` `" " = t e ^(t) - int 1* e ^(t) dt + C ` ` " " = t e ^(t) - e ^(t) +C = e^(t) (t -1 ) +C ` `" "= e ^(tan x ) (tanx - 1 ) +C `. Hence, `y = (tanx - 1 ) + Ce ^(-tan x )` is the required solution. |
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| 22. |
`(dy)/(dx) + 2y = e ^(-2x) sinx `, given that `y=0` when ` x = 0 ` |
| Answer» Correct Answer - `y = sinx ` | |
| 23. |
` x dy - ( y+ 2x ^(2)) dx = 0 ` |
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Answer» Correct Answer - `y = 2x ^(2) + Cx ` `(dy)/(dx) = (y + 2 x ^(2))/(x) rArr (dy)/(dx) - (1)/(x) *y = 2x ` |
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| 24. |
The solution of the DE ` (dy)/(dx) + y log y cot x = 0 ` isA. `cos x log y =C `B. `sin x log y= C `C. ` log y = C sin x `D. none of these |
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Answer» Correct Answer - B ` (1)/( y log y ) dy + cot xdx = 0 rArr (dy)/(y log y) + int cot x dx = log C ` `rArr log ( log y ) + log sinx = log C rArr log (sin x + log y ) =log C ` `rArr (sin x ) log y = C `. |
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| 25. |
The solution of the DE `x (dy)/(dx) = cot y ` isA. ` x cos y = C `B. `x tan y = C `C. `x sec y = C `D. none of these |
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Answer» Correct Answer - A `int tan y dy = int (1)/(x) dx rArr log x = log sec y +log C.` ` rArr x = C sec y rArr x cos y =C`. |
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| 26. |
The solution of the DE ` x dy + y dx = 0 ` isA. `x+y = C `B. `xy = C `C. ` log (x + y )= C `D. none of these |
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Answer» Correct Answer - B ` xdy = y dx = 0 rArr int (1)/(y) dx = - int (1)/(x) dx rArr log y = - log x + logC`. ` therefore log xy =log C rArr xy = C`. |
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| 27. |
Solve`[(e^(-2sqrt(x)))/(sqrt(x))-y/(sqrt(x))](dx)/(dy)=1(x!=0` |
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Answer» The given differential equation may be written as ` (dy )/(dx) = - (1)/(sqrtx) * y + ( e ^( -2sqrt x ))/( sqrtx )` `rArr (dy)/(dx) + (1)/(sqrt x) * y = ( e ^(-2sqrt x))/( sqrt x )" "` ... (i) This is of the form ` (dy)/(dx) + Py= Q`, where ` P = (1)/(sqrtx ) and Q = ( e ^(-2sqrtx ))/( sqrt x )` `IF = e ^(int Pdx ) = e ^(int (1)/(sqrtx) dx ) = e ^(2sqrt x ) ` So, the solution of the given differential equation is given by `y xx IF = int (Q xx IF ) dx + C ` i.e., `y xx e ^( 2sqrtx) = int ( e ^(-2sqrtx))/(sqrtx ) xx e ^(2sqrt x ) dx + C ` `" " = int (1)/(sqrtx)dx + C = 2 sqrtx + C`. Hence, ` y e^(2sqrtx) = 2 sqrtx + C ` is the required solution. ` therefore y = (1)/(x) - cot x + (C)/(x sin x ) ` is the required solution. |
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| 28. |
The solution of the DE ` (e ^(x) + 1) y dy= ( y + 1) e ^(x) dx` isA. ` e ^(y) = C (e ^(x) + 1) ( y + 1 )`B. `e^(y)= e ^(x) + y + 1 `C. ` y = (e ^(x) + 1 ) ( y + 1)`D. none of these |
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Answer» Correct Answer - A `int ( e ^(x)) /(( e ^(x) + 1 )) dx = int (( y )/( y + 1 )) dy rArr log ( e ^(x) + 1 ) = int (1 - (1)/( y + 1 )) dy ` `rArr y - log ( y + 1 ) = log ( e ^(x) + 1 ) + log C rArr y = log [C( e ^(x) + 1 ) ( y + 1) ]` ` rArr e ^(y) = C ( e ^(x) + 1) ( y + 1 )` |
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| 29. |
Solve the differential equation : `x(dy)/(dx)+y-x+xycotx=0`, `x ne 0`. |
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Answer» The given differential may be written as ` x (dy)/(dx) + (1 + x cot x )y = x ` `rArr (dy)/(dx) + ((1)/(x) + cot x ) y = 1 ` This is of the form ` (dy)/(dx) + Py = Q`, where ` P = ((1)/(x) + cot x ) and Q = 1 ` Thus, the given differential equation is linear. ` IF = e ^(int Pdx) = e ^( int ((1)/(x) + cot x ) dx) = e ^(log x + log sin x) ` `" " = e ^( log (x sinx )) = x sinx ` So, the solution of the given differential equation is given by `y xx (IF) = int (Q xx IF) dx + C`, i.e. , `y (x sin x) = int ( 1 xx (x sin x ) dx +C ` ` " " = int underset("I")( x )underset("II") (sin) x dx +C` ` " " = x ( - cos x ) - int 1 *( - cos x ) dx + C ` ` " " = - xcos x + int cos x dx + C ` ` " " = - x cos x + sin x +C`. ` therefore y = (1)/(x) - cot x + (C)/( x sin x ) ` is the required solution. |
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| 30. |
Find the general solution of the linear differential equation `(x+y) dy/dx=1.` |
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Answer» Correct Answer - ` (x+ y + 1 )e ^(-y) =C ` ` ( x+ y ) (dy)/(dx) = 1 rArr (dx)/(dy) = x + y ` `therefore (dx)/(dy) - x = y ` ` IF = e ^(-int dy ) = e ^(-y)`. ` therefore x xx e ^(-y) = int underset ("I")y underset ("II") e ^(-y) dy+C`. |
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| 31. |
The solution of the `DE (dy)/(dx) = 2 ^( x + y )` isA. ` 2 ^(x ) + 2^(y) = C `B. ` 2 ^(x) + 2 ^(-y) = C `C. `2 ^(x) - 2 ^(-y) = C `D. none of these |
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Answer» Correct Answer - B We have ` 2 ^(x) dx = 2 ^(-y) dy rArr int 2^(x) dx = int 2 ^(-y) dy`. ` therefore 2 ^(x) log 2 = - 2 ^(-y) log 2 = log C rArr 2 ^(x) + 2 ^(-y) =C `, where ` (log C )/( log 2) = C.` |
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| 32. |
The solution of the DE ` (dy)/(dx) = (( 1+y ^(2)))/((1 + x ^(2))) ` isA. ` (y + x) = C ( 1 - yx)`B. ` (y - x ) = C (1 + y x )`C. `y = (1 + x) C `D. none of these |
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Answer» Correct Answer - B `int ( dy ) /((1 + y^(2))) = int (dx)/((1 +x ^(2))) rArr tan ^(-1) y = tan ^(-1) x + C_1 ` ` rArr tan ^(-1) y - tan ^(-1) x = C_1 rArr tan ^(-1) (( y - x )/( 1+ yx )) = C_1 rArr ( y - x )/((1 + y x ) ) = tan C_1 = C`. |
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| 33. |
The solution of the DE ` (dy)/(dx) = (1 - cos x )/( 1+ cos x )` isA. `y = 2 tan ""(x)/(2) - x + C `B. ` y tan ""(x)/(2) - 2x + C `C. ` y = tan x - x + C `D. none of these |
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Answer» Correct Answer - A ` (dy)/(x) = (1 - cos x )/( 1+ cos x) = (2 sin ^(2) (x//2))/( 2 cos ^(2) (x //2)) = tan ^(2) ""(x)/(2) = ( sec ^(2) ""(x)/(2) - 1 )` ` rArr int dy = int ( sec ^(2) ""(x)/(2) - 1 ) dx rArr y = 2 tan ""(x)/(2) - x + C`. |
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| 34. |
The solution of the DE `cos x (1 + cosy )dx - sin y (1+ sin x ) dy=0 ` isA. `1 + sin x cos y = C `B. ` (1 + sin x ) (1 + cos y ) = C `C. `sinx cos y + cos x = C `D. none of these |
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Answer» Correct Answer - B `int (cos x )/((1 +sin x )) dx - int (sin y )/((1+ cos y )) dy = log C` ` rArr log (1 + sin x ) + log (1 + cos y ) = log C rArr (1 + sin x ) (1 + cos y ) =C`. |
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| 35. |
`y dx+(x-y^2)dy=0` |
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Answer» Correct Answer - ` x = ( y ^(2))/( 3) + (C )/(y)` `y dx = ( y^(2) - x ) dy rArr (dx)/(dy) = ( y ^(2) - x )/( y )` ` therefore (dx)/(dy) - (1)/(y) * x = y`. |
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| 36. |
The solution of the DE `x cos y dy = (x e ^(x) log x + e ^(x)) dx ` isA. `sin y = e ^(x) + log x + C `B. ` sin y - e ^(x) +log x = C `C. `sin y = e ^(x) (log x ) + C `D. none of these |
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Answer» Correct Answer - C `int cos y dy = int e ^(x) (log x + (1)/(x) ) dx rArr sin y = e ^(x) log x + C`. |
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| 37. |
`x dy/dx -y= log x,` given that `y = 0` when `x = 1.` |
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Answer» Correct Answer - `y = x - 1 - log x ` `IF = e ^( -int (1)/(x) dx ) = e ^(-log x ) = e^( log ( 1//x)) = (1)/(x)` ` y xx (1)/(x) = int ( underset ("I" ) log x *underset ("II") ((1)/(x^(2)))) dx +C`. |
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| 38. |
Solve the differential equation ` x (dy )/(dx) - y = 2 x^(3), x gt 0` |
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Answer» The given differential equation may be written as ` (dy)/(dx) + (( - 1 )/( x )) y = 2x ^(2)" " ` ... (i) This is of the form, ` (dy)/(dx) + Py = Q`, where `P = (-1)/(x) and Q = 2x ^(2)` Thus, the given diffential equation is linear. IF ` = e ^( int Pdx) = e ^( int - (1)/(x) dx ) = e ^(-log x ) = e ^( log ((1)/(x))) = (1)/(x)` So, the required solution is given by ` y xx IF = int {Q xx IF} dx + C` , where C is an arbitrary constant, i.e, ` y xx (1)/(x) = int ( 2x ^(2) xx (1)/(x)) dx + C = 2 int x dx + C = x ^(2) + C ` `rArr y = x ^(3) + Cx ` Hence, the required solution is `y = x ^(3) + Cx` |
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| 39. |
Find the particular solution of the differential equation ` (1 - x ^(2) ) (dy)/(dx) - xy = x ^(2)`, given that `y = 2 ` when `x = 0`. |
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Answer» The given differential equation may be written as ` (dy)/(dx) - (x)/(( 1- x ^(2))) * y = (x ^(2))/((1- x ^(2)))" "`... (i) This is of the form ` (dy)/(dx) + Py = Q`, where `P = (-x)/((1-x ^(2))) and Q = (x ^(2))/((1 - x ^(2)))` Thus, the given differential equation is linear. `IF= e ^(int Pdx) = e ^(int (-x)/((1- x ^(2))) dx ) = e ^((1)/(2) int (-2x)/((1- x ^(2))) dx ) = e ^((1)/(2) log ( 1 - x ^(2)))` ` " " = e ^(log sqrt (1 - x^(2))) = sqrt (1- x ^(2))` So, its solution is given by ` y xx IF = int (Qxx IF ) dx + C`, i.e., `y xx sqrt (1- x ^(2)) = int (x^(2))/(( 1- x ^(2))) xx sqrt (1 - x ^(2)) dx + C ` ` " " = int (x^(2))/(sqrt (1- x ^(2))) dx + C ` ` " " int ({ - ( 1 -x ^(2)) + 1})/( sqrt (1 - x ^(2))) dx + C ` ` " " = - int sqrt (1 - x ^(2)) dx + int (1)/(sqrt (1 - x ^(2))) dx +C ` `= -{ (x sqrt (1- x ^(2)))/(2) + (1)/(2) sin ^(-1) x } + sin ^(-1) x + C ` ` " " = (-x sqrt (1- x ^(2)))/( 2 ) + (1)/(2) sin ^(-1) x +C . ` ` therefore y = (-x)/(2) + (sin ^(-1)x )/(2 sqrt (1- x ^(2))) + (C)/( sqrt (1- x ^(2)))" " ` ... (ii) It is being given that when ` x = 0 `, then `y = 2 ` Putting ` x=0 and y =2 `in (ii), we get C = 2 Hence , `y = (-x)/(2) + (sin ^(-1)x)/(2sqrt( 1- x ^(2))) + (2)/(sqrt (1 - x ^(2)))` is the required solution. |
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| 40. |
An equation relating to the stability of an aeroplane is given by ` (dv)/(dt) = g cos alpha - kv`, where ` v` is the velocity and ` g, alpha, k` are constants. Find an expression for the velocity if `v = 0` at ` t =0` |
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Answer» The given differential equation is ` (dv)/(dt) + kv= g cos alpha " "` ... (i) This is of the form ` (dv)/(dt) + Pv = Q`, where `P = k and Q = g cos alpha `. Thus, the given equation is linear. `IF = e ^(int Pdx) = e ^(int k dt ) = e^(kt )` So, the solution of the given differential equation is `vxx IF = int {Q xx IF}dt + C`, i.e., `ve^(kt) = int (gcos alpha ) e^(kt) dt + C` `" " = ((g cos alpha ) e^(kt))/( k ) +C" " `... (ii) Now, it is given that ` v =0` when `t = 0` Putting `t=0 and v=0 `in (ii), we get `C = (-g cos alpha)/( k)` ` therefore ve^(kt) = ((g cos alpha ) e^(kt))/(k) - (gcos alpha )/( k)` `rArr v = (1)/(k) (g cos alpha )( 1 - e^(kt))`, which is the required expression. SOLUTION OF `(dx)/(dy) + Px =Q` Working Rule for Solving ` (dx)/(dy) +Px = Q ` (i) Find If = ` e ^(int Pdy)` (ii) The solution is given by ` x xx IF = int {Q xx IF}dy +C`. |
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| 41. |
Solve the following differential equation :`(("x"^2-1)"dy")/("dx")+2"x y"=2/(("x"^2-1))` |
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Answer» The given differential equation may be written as `(dy)/(dx) + { (2x)/(( x ^(2) - 1 ))} y = (2)/(( x ^(2) - 1 ) ^(2)) " " ` ... (i) This is of the form ` (dy)/(dx) + Py = Q, ` where `P = (2x)/((x ^(2) - 1 )) and Q = (2)/( ( x^(2) - 1 ) ^(2)) ` Thus, the given differential equatiion is linear. ` IF = e ^(int Pdx) = e ^(int (2x)/(( x ^(2)-1)) dx) = e ^( log (x^(2) -1) ) = ( x ^(2) - 1 )` So, the required solution is given by `yxx IF = int {Q xx IF } dx + C ` i.e., `y xx (x ^(2) -1 ) = int { (2)/(( x ^(2) - 1 ) ^(2)) xx ( x ^(2) - 1 ) } dx +C ` ` " " = 2 int (dx )/(( x ^(2) - 1 )) + C ` ` " " = 2 int (1)/(2) {(1)/(( x - 1 )) - (1)/(( x + 1 )) } dx + C ` [ by partial fraction ] ` " " = log |( x - 1 )/( x+ 1 )| + C .` Hence, ` y (x ^(2) - 1 ) = log |(x - 1 )/(x + 1)| + C` is the required solution. |
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| 42. |
`(y+3x^2)(d x)/(d y)=x` |
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Answer» Correct Answer - ` y = 3x ^(2) + Cx ` `(dy)/(dx) = (y + 3x ^(2))/(x) rArr (dy)/(dx) - (1)/(x)* y = 3 x ` |
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| 43. |
Findthe equation of a curve passing through the point (0, 1). If the slope of thetangent to the curve at any point (x, y) is equal to the sum of the xcoordinate (abscissa) and the product of the x coordinate and y coordinate(ordinate) of t |
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Answer» We know that the slope of the tangent to the curve is ` (dy)/(dx)` ` therefore (dy)/(dx) = x + xy rArr (dy)/(dx) - xy = x " " `... (i) This is of the form ` (dy)/(dx) + Py = Q`, where ` P = -x and Q = x `. So, the given differential equation is linear. `IF= e ^(int Pdx) = e ^(int - x dx) = e^((-x ^(2))/( 2 ))` Hence, the solution of the given differential equation is given by `y xx IF = int (Q xx IF )dx +C`, i.e., `y xx e ^((-x^(2))/(2)) = int x e^((-x^(2))/( 2))dx + C ` ` = int e ^(-t) dt +C `, where ` (x^(2))/(2) = t ` ` " " = - e ^(-t) + C = - e ^((-x^(2))/( 2 )) +C ` `therefore " " y= -1 +Ce ^((x ^(2))/( 2 ))" " `... (ii) We have to find a curve satisfying (ii) and passing through (0, 1). Putting ` x = 0 and y = 1 `in (ii), we get ` C = 2`. Hence, ` y = -1 + 2e^((x ^(2))/( 2)) ` is the equation of the required curve. |
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| 44. |
Solve ` (1 + x ^(2)) (dy )/(dx) + y = e ^( tan ^(-1) x )`. |
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Answer» The given differential equation may be written as ` (dy)/(dx) + (1)/(( 1 + x ^(2)) ) * y = ( e ^( tan ^(-1) x )) /( ( 1+ x ^(2))) ` This is of the formm ` (dy )/(dx) + P y = Q ` , where `P = (1)/( ( 1 + x ^(2))) and Q = (e ^(tan ^(-1) x ))/( (1 + x ^(2)))` Thus, the given equation is linear. `IF = e ^(int Pdx) = e ^( int (1)/(( 1 + x ^(2))) dx) = e ^(tan ^(-1) x )`. So, the required solution is given by `y xx IF = int |( Q xx ( IF) | dx + C`, i.e, `y xx e ^( tan ^(-1) x ) = int { ( e ^(tan ^(-1)x))/( (1 + x ^(2))) xx e ^(tan ^(-1)x)} dx + C ` ` " " = int ( e ^(2tan ^(-1) x ))/( (1 + x ^(2))) dx + C ` ` " " = int e ^(2t)dt + C`, where ` tan ^(-1)x = t ` ` = (1)/(2) e ^( 2t ) + C = (1)/(2) e ^(2tan^(-1)x) + C`. Hence, `y = (1)/(2) e^(tan ^(-1) x ) + C e^(-tan ^(-1) x ) ` is the required solution. |
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| 45. |
Findthe equation of a curve passing through the point (0, 2) given that the sumof the coordinates of any point on the curve exceeds the magnitude of theslope of the tangent to the curve at that point by 5. |
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Answer» Correct Answer - `y = 4 - x - 2e ^(x)` ` (x +y) - (dy )/(dx) = 5`. |
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| 46. |
` (dy)/(dx) + (1)/(x) * y = x ^(2)` |
| Answer» Correct Answer - `y= (x ^(3))/( 4) + (C)/(x)` | |
| 47. |
Solve ` (x log x ) ( dy)/(dx) + y = (2)/(x) log x `. |
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Answer» The given differential equation may be written as ` (dy)/(dx) + (1)/( (x logx )) * y = ( 2)/(x^(2)) `. This is of the form ` (dy)/(dx) + Py =Q, ` where `P = (1)/((x log x )) and Q = (2)/(x ^(2)) ` Thus, the given differential equation is linear. ` IF = e ^(int Pdx) = e ^( int (1)/(x log x) dx ) = e ^( int (1)/(t) dt)`, where ` log x = t ` `" " = e ^( log t) = t = log x ` So, the solution of the given differential equation is ` y xx IF = int {Q xx (IF)}dx + C`, i.e., `y (log x ) = int(( 2)/( x^(2)) log x )dx +C ` `" " = 2 int ( log x ) * (1)/(x ^(2)) dx+ C ` `" " = 2 [ (logx ) (- (1)/(x)) - int (1)/(x) * (- (1)/(x)) dx ] + C`. `" " ` [ integrating by parts] `" " = ( - 2log x ) /(x) - (2)/(x) + C` Hence, `y (log x ) = (-2)/(x) (log x + 1 ) +C ` is the required solution. |
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| 48. |
` (x log x ) (dy)/(dx) + y = 2 log x ` |
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Answer» Correct Answer - `y(logx) = (log x ) ^(2) +C ` ` (dy)/(dx) + (1)/(x log x ) * y = (2)/(x)` ` IF = e ^(int (1)/(x log x )) dx = e ^(int (1)/(t) dt) = e ^( log t ) = t = log x ` |
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| 49. |
`(dy)/(dx) + 2 y tan x = sin x ` |
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Answer» Correct Answer - `y = cos x + C cos ^(2) x` `IF = e^ ( int 2 tan x dx ) = e ^(-2 log cos x ) = e ^(log (cos x ) ^(-2)) = (cos x) ^(-2) = (1)/(cos ^(2) x)` ` y xx (1)/(cos^(2) x) = int { sinx xx (1)/(cos ^(2) x )} dx +C = int sec x tan x dx +C = sec x +C ` ` therefore y =cos x + C cos ^(2)x ` |
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| 50. |
` x (dy)/(dx) + y = x logx ` |
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Answer» Correct Answer - ` 4xy = 2 x^(2) log x - x ^(2) +C ` |
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