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51.

The solution of `(dy)/(dx) + y = e^(x)` isA. ` 2y = e^(2x)+C`B. `2ye^(x)=e^(2)+C`C. `2ye^(x)=e^(2x)+C`D. `2ye^(2x)=2e^(x)+C`

Answer» Correct Answer - d
Given differential equation is `(dy)/(dx) +y = e^(x)`
This is linear differential equation
` :. IF = e^(int P dx) = e^(int 1 dx ) = e^(x)` ,
Now, solution is ` ye^(x) = int e^(2x) dx +C/2 `
` rArr ye^(x) = (e^(2x))/2 + C/2 rArr 2ye^(x) = e^(2x) +C`
Discussion
52.

The solution of the differential equation `((x+2y^3)dy)/(dx)=y`is(a)`( b ) (c) (d) x/( e )(( f ) (g) y^(( h )2( i ))( j ))( k ) (l)=y+c (m)`(n)(b) `( o ) (p) (q) x/( r ) y (s) (t)=( u ) y^(( v )2( w ))( x )+c (y)`(z)(c)`( d ) (e) (f)(( g ) (h) x^(( i )2( j ))( k ))/( l ) y (m) (n)=( o ) y^(( p )2( q ))( r )+c (s)`(t)(d) `( u ) (v) (w) y/( x ) x (y) (z)=( a a ) x^(( b b )2( c c ))( d d )+c (ee)`(ff)A. `x=y^(3) +Cy`B. `x = y^(3)+2Cy`C. ` x = 2y^(3)+Cy`D. ` x=3y^(3)+Cy`

Answer» Correct Answer - a
The Given differential equation can be written as
` (dx)/(dy ) + (-1/y) x = 2y^(2) , y != 0 `
This is a linear differential equation of the form
` (dx)/(dy) +Px =Q `
Here, `P = (-1)/y and Q = 2y^(2)`
` :. IF = e^( intPdy ) = e^(int (-1)/y dy) = e^(-logy ) = y^(-1) = y^(-1) = 1/y `
The general solution of the given differential equation is given by
` x IF = int (Q xx IF) dy + C`
` rArr x(1/y) = int 2y^(2) *1/y dy +C`
` rArr x/y = y^(2) +C rArr x = y^(3) + Cy`
Discussion
53.

To reduce the differential equation `(dy)/(dx) + P (x)*y = Q (x) * y^(n)` to the linear form , the substitution isA. `v = 1/(y^(n))`B. ` v = 1/(y^(n-1))`C. ` v = y^(n)`D. `v = y^(n-1)`

Answer» Correct Answer - a
The Given differential equation is
` (dy)/(dx) + P (x)* y = Q (x) * y^(n)`
` rArr 1/(y^(n)) * (dy)/(dx) + y^(-n+1) P y = Q (x)* y^(n)`
Put ` 1/(y^(n-1))= v `
` rArr (-n +1) y^(-n)(dy)/(dx) =(dv)/(dx)`
` :. 1/(-n+1) * (dv)/(dx) +P (x) * v = Q (x)`
` rArr (dv)/(dx) +(1-n) P (x) * v = (1-n) Q (x)`
which is linear differential equation .
Hence , required substitiution is ` v = 1/(y^(n-1))`
Discussion
54.

The solution of the differential equation `(dy)/(dx) = e^(3x-2y) +x^(2)e^(-2y)`,isA. `e^(2y)=e^(3x)+x^(3)+c`B. `e^(2y)=e^(3x)-x^(3)+c`C. `(e^(2y))/(2)=(1)/(3)(e^(3x-x3))+c`D. `(e^(2y))/(2)=(1)/(3)(e^(3x+x3))+c`

Answer» Correct Answer - D
Discussion
55.

An integrating factor of the differential equation `(1+y+x^(2)y) dx + ( x +x^(3))dy = 0` isA. log xB. xC. `e^(x)`D. `1/x`

Answer» Correct Answer - a
Given , `(1+y+x^(2)y) dx + ( x + x^(3) )dy = 0 ` ,
` rArr (dy)/(dx) = - (1+y+x^(2)y)/(x+x^(3))`
` rArr (dy)/(dx) +y/x = - 1/(x(1+x^(2)))`
which is linear differential equation
` :. IF = e^(int 1/x dx) = e^(logx)= x`
Discussion
56.

`2x^(x+2y) dx -3dy =0`A. `4e^(x)+3e^(-2y)=c`B. `4e^(x)-3e^(-2y)=c`C. `4e^(x)+3e^(2y)=c`D. `4e^(x)-3e^(2y)=c`

Answer» Correct Answer - A
Discussion
57.

The solution of the diffferential equation ` x (dy)/(dx ) + 2y = x^(2) ` isA. ` y = (x^(2)+C)/(4x^(2))`B. `y = (x^(2))/4+C`C. `y= (x^(4)+C)/(x^(2))`D. ` y = (x^(4)+C)/(4x^(2))`

Answer» Correct Answer - b
Given , `(dy)/(dx) +2/x y = x `
This is linear differential equation
` :. " Integrating factor " =e^(int 2/x dx) = x^(2) `
` :. "Required solution is "`
` y * x^(@) = int x^(3) dx + C/4 = (x^(4))/4 +C/4 `
` rArr y = (x^(4)+C)/(4x^(2))`
Discussion
58.

Solution of differential equation `e^((dy)/(dx))=x` isA. `xlog|x|-1-y=c`B. `xlog|x|-1+y=c`C. `x(log|x|-1)+y=c`D. `x(log|x|-1)-y=c`

Answer» Correct Answer - D
Discussion
59.

The solution of differential equation ` (x^(2)-1) (dy)/(dx) + 2xy = 1/(x^(2)-1)` isA. ` y(x^(2)-1)=1/2 log |(x-1)/(x+1)|+C`B. ` y (x^(2)+1)=1/2 log |(x-1)/(x+1)|+C`C. ` y (x^(2)+1)=1/3 log |(x-1)/(x+1)|+C`D. `y ( x^(2)-1)=1/3 log |(x-1)/(x+1)|+C`

Answer» Correct Answer - d
The given differential equation is
` (x^(2)-1) (dy)/(dx) +2xy = 1/(x^(2)-1)`
` rArr (dy)/(dx) +(2x)/(x^(2)-1) y = 1/(x^(2)-1)^(2)`
This is a linear differential equation is
Here, `P = (2x)/(x^(2)-1) and Q = 1/(x^(2)-1)^(2)`
` :. IF = e^(int Pdx) = e^(int (2x)/(x^(2)-1)dx) = e^(log (x^(2)-1))= (x^(2)-1))`
The general solution of the given differential equation is
`y* IF int Q xx IF dx +C`
` rArr y(x^(2)-1) = int 1/(x^(2)-1)dx +C`
`rArr y (x^(2)-1) = 1/2 log |(x-1)/(x+1)| +C`
Discussion
60.

The solution of the differential equation `(1+y^(2)) dx = (tan^(-1) y - x) dy` isA. ` x = tan ^(-1)y -1+Ce^(-tan^(-1)y)`B. ` y = tan^(-1)y+1 +Ce^(-tan^(-1)y)`C. ` x = tan^(-1)y +Ce^(-tan^(-1))y`D. None of the above

Answer» Correct Answer - a
The Given differential equation is
`(1+y^(2)) dx = (tan^(-1) y - x) dy`
This equation can be rewritten as
` (dx)/(dy) + x/(1+y^(2)) = (tan^(-1)y)/(1+y^(2))`
This is of the form ,
`(dx)/(dy) +Px = Q`
Here, ` P = 1/(1+y^(2)) and Q = (tan^(-1)y)/(1+y^(2))`
` :. IF = e^(int 1/(1+y^(2))dy) = e^(tan-1y)`
Therefore , required solution is
` xe^(tan-1y) = int e^(tan^(-1)) * (tan^(-1)y)/(1+y^(2)) dy +C`
` rArr xe^(tan-1y) = (tan ^(-1)y-1)e^(tan^(-1)y) +C`
` rArr x = tan^(-1) y - 1 +Ce ^(-tan ^(-1)y)`
Discussion
61.

Solution of differential equation `(dy)/(dx)=3^(x+y)` isA. `3^(x)-3^(-y)=c`B. `3^(x)+3^(-y)=c`C. `3^(x)-3^(y)=c`D. `3^(x)-3^(y)=c`

Answer» Correct Answer - B
Discussion
62.

Solution of the differential equation `x-tan^(-1)y-1c e^(-tan^(-1)y)` isA. `y=(c-x)e^(x)`B. `y=(c-x)e^(-x)`C. `y=(c+x)e^(x)`D. `y=(c+x)e^(-x)`

Answer» Correct Answer - D
Discussion
63.

Solve the following differential equation: `(1+y^2)tan^(-1)dx+2y(1+x^2)dy=0`A. `(1)/(2)tan^(-1)x+log|1+y^(2)|=c`B. `(1)/(2)tan^(-1)x-log|1+y^(2)|=c`C. `(1)/(2)tan^(-1)x+log|1+y^(2)|=c`D. `(1)/(2)tan^(-1)x-log|1+y^(2)|=c`

Answer» Correct Answer - C
Discussion
64.

Solution of the differential equation `(1+x^(2)) dy + 2xy dx = cot x dx ` isA. ` y = log |sin x | (1+x^(2))+C(1+x^(2))^(-1)`B. ` y = log |sin x|(1+x^(2))^(-1)+C(1+x^(2))`C. ` y = log |sin x | (1+x^(2))+C(1+x)`D. ` y = log |sin x| (1+x^(2))^(-1)`

Answer» Correct Answer - c
Given , `(1+x^(2))dy +2xy dx = cot x dx `
` rArr (1+x^(2)) dy = dx (cot x - 2xy ) `
` rArr (dy)/(dx) = (cot x - 2xy)/(1+x^(2))`
` rArr (dy)/(dx) +(2xy)/(1+x^(2))= (cot x)/(1+x^(2))`
Or comparing with the form `(dy)/(dx) +Py = Q` we get
` :. P = (2x)/(1+x^(2))and Q = (cot x)/(1+x^(2))`
` :. IF = e^(int Pdx) = e^(int (2x)/(1+x^(2) dx)) = e^(log(1+ x^(2))= 1+x^(2)`
The General solution of the given differential equation given by
`y * IF int Q xx IF dx +C`
` rArr (1+x)^(2) y = int( (1+x^(2))(cotx)/((1+x^(2)))) dx +C`
` rArr (1+x^(2)) y = log |sin x| +C`
` rArr y = log | sin x | (1+x^(2))^(-1)+C (1+x^(2))^(-1)`
Discussion
65.

solution of the equation `(dy)/(dx)+1/x tany=1/x^2 tanysiny` isA. ` 2x= (1-2Cx^(2))sin y`B. `x=(1-2 Cx^(2)) sin y`C. `2x = (1+2Cx^(2))sin y`D. None of these

Answer» Correct Answer - d
The given differential equation can be rewritten as ` cot y "cosec" y (dy)/(dx) +1/x "cosec" y = 1/(x^(2))`
On putting ` v = - "cosec" y and (dv)/(dx) = "cosec" y cot y (dy)/(dx),`
we get ` (dv)/(dx) -1/x v = 1/(x^(2))`
which is linear differential equation in v.
Here, `P = -1/x and Q = 1/(x^(2))`
` :. IF = e^(int Pdx)= e^(int - 1/xdx)= e^(-logx)=1/x`
Required solution is
` v = 1/x = int 1/x * 1/(x^(2))dx +C = -1/(2x^(2))+C`
` rArr 2x = (1- 2Cx^(2) ) sin y `
Discussion
66.

The solution of the differential equation `(1+y^(2)) dx = (tan^(-1) y - x) dy` isA. `x+tan^(-1)y+1c e^(-tan^(-1)y)`B. `x-tan^(-1)y+1c e^(-tan^(-1)y)`C. `x+tan^(-1)y-1c e^(-tan^(-1)y)`D. `x-tan^(-1)y-1c e^(-tan^(-1)y)`

Answer» Correct Answer - B
Discussion
67.

The solution of the differential equation ` (dy)/(dx) - (tany)/x = (tan y sin y)/(x^(2))` isA. `x/(siny) + log x = C`B. `y/(sin x)+log x =C`C. `log x +x = C`D. `log x +y = C`

Answer» Correct Answer - b
Given , ` (dy)/(dx) - (tany)/x = (tany sin y )/(x^(2))`
` rArr coty " cosec " y (dy)/(dx) - ("cosec "y)/x = 1/(x^(2))`
Put ` - " cosec " y = t`
` rArr cot y " cosec " y (dy)/(dx) = (dt)/(dx)`
Eq. (I ) reduces to , ` (dt)/(dx) +t/x = 1/(x^(2))`
` IF = e^(int P dx) = e^(int 1/x dx)= x`
` :. ` Required solution is given by
` tx = int x * 1/(x^(2)) dx- C`
`rArr - "cosec" y * x = log x = C`
` rArr x/(sin y) + log x = C `
Discussion
68.

Solution of the differential equation `(dy)/(dx)tany=sin(x+y)+sin(x-y)` isA. `secy+2cosx=c`B. `secy-2cosx=c`C. `cosy-2sinx=c`D. `tany-2secx=c`

Answer» Correct Answer - A
Discussion
69.

The differential equation obtained by eliminating a and b from `y = ae^(bx)` isA. `y(d^(2)y)/(dx^(2))-((dy)/(dx))^(2)=0`B. `y(d^(2)y)/(dx^(2))+((dy)/(dx))^(2)=0`C. `(d^(2)y)/(dx^(2))-((dy)/(dx))^(2)=0`D. `(d^(2)y)/(dx^(2))+((dy)/(dx))^(2)=0`

Answer» Correct Answer - A
Discussion
70.

The differential equation ` (e^(x)+1)y dy = (y+1) e^(x) dx ` has the solutionA. `(y-1)(e^(x)-1) = Ce^(y)`B. ` (y-1)(e^(x)+1) = Ce^(y)`C. ` (y+1) (e^(x)-1)=Ce^(y)`D. ` (y+1)(e^(x)+1)= Ce^(y)`

Answer» Correct Answer - c
Given differential equation can be rewritten as
` (y dy)/(y+1) = (e^(x)dx)/(e^(x)+1)`
` rArr (1- 1/(y+1))dy = (e^(x))/(e^(x)+1)dx`
` rArr y - log (y+1) = log (e^(x)+1) - log C`
[ integrating both sides ]
` rArr y = log. ((e^(x)+1)(y+1))/C` ,
` rArr (e^(x)+1) (y+1) = Ce^(y)`
Discussion
71.

The solution of the differential equation `(dy)/(dx) = e^(3x-2y) +x^(2)e^(-2y)`,isA. ` e^(2y) +e^(3x) +x^(3) +C`B. ` 1/2 e^(2y) =1/3 (e^(3x)+x^(3))+C`C. ` 1/2 e^(2y) = 1/3 (e^(3x)+x^(3))+C `D. ` e ^(2y)= e^(3x) + x^(2) +C`

Answer» Correct Answer - d
Given differential equation can be rewritten as ,
` e^(2y)dy = (e^(3x)+x^(2))dx`
On integrating , we get
` rArr (e^(2y))/2 = (e^(3x))/3 +x^(3)/3 + C`
Discussion
72.

The solution of the differential equation ` y - x ( dy)/(dx) = a ( y^(2) + dy/dx)` isA. `y = C (x+a)(1-ay)`B. `y = C( x+a) (1+ay)`C. `y = C(x-a)(1+ay)`D. None of these

Answer» Correct Answer - d
Given , ` y - x (dy)/(dx) = a (y^(2) +dy/dx)`
` rArr (dy)/(dx) (a+x) = y - ay^(2) `
On integrating both sides , we get
` rArr int (1/y +a/(1-ay))dy = int (dx)/(a+x)`
` rArr log y - log (1-ay) = log (a+x) +log C`
` rArr log y = log (1- ay) (a+x)C`
` rArr y = C ( 1-ay ) (a+x)`
Discussion
73.

The differential equation of the family of straight lines whose slope is equal to y - intercept ,isA. `a(d^(2)y)/(dx^(2))+((dy)/(dx))^(3)=0`B. ` 2a (d^(2)y)/(dx^(2))+((dy)/(dx))^(3)=0`C. ` 2a (d^(2)y)/(dx^(2))- ((dy)/(dx))^(3)`D. ` a (d^(2)y)/(dx^(2))-((dy)/(dx))^(3)=0`

Answer» Correct Answer - b
Equation of the family of such parabola is
` ( y - k)^(2) = 4a (x-h) " "` …(i)
where h and k are arbitrary constants ltbRgt On differetiating w.r.t , we get
`(y-k) (dy)/(dx) = 2a " " ` (ii)
Again , on differentiating , we get
` (y - k) (d^(2)y)/(dx^(2)) + ((dy)/(dx))^(2)=0" "`…(iii)
On putting hte value of `(y-k) ` from Eq. (ii) in Eq. (iii) , we get
` 2a (d^(2)y)/(dx^(2)) + ((dy)/(x))^(3) = 0 ,`
which is the required differential equation
Discussion
74.

The solution of the differential equation `(dy)/(dx) = e^(y+x) +e^(y-x)` isA. ` e^(-y) = e^(x) -e^(-x)+C`B. ` e^(-y) = e^(-x)-e^(x) +C`C. ` e^(-y) = e^(x) +e^(-x)+C`D. ` e^(-y) +e^(x) +e^(-x) =C`

Answer» Correct Answer - b
Given differential equation is
` (dy)/(dx) = e^(y+x) +e^(y-x)`
On integrating both sides , we get
` rArr int e^(-y) dy = int (e^(x)+e^(-x))dx`
` rArr -e^(-y) = e^(x) -e^(-x) -C`
` rArr e^(-y) = e^(-x) -e^(x) +C`
Discussion
75.

Finda particular solution of the differential equation`(x" "" "y)" "(dx" "+" "dy)" "=" "dx" "" "dy`, given that `y" "=" "1`, when `x" "=" "0`. (Hint: put `x" "" "y" "=" "t`).A. ` log |x-y|= x+y +1`B. ` log |x-y|=x - y -1`C. ` log |x-y|= x+y -1`D. None of these

Answer» Correct Answer - a
Given , differential equation is
` (x-y) (dx+dy) = dx - dy`
` rArr dx + dy = (dx-dy)/(x-y)`
On integrating both sides , we get
` int (dx + dy) = int (dx-dy)/(x-y) +C`
Let `x - y = t rArr dx - dy = dt`
` :. int dx + dy -C = int (dx-dy)/(x-y) = int (dt)/t = log t = = log |x-y|`
` rArr x+ y = log + C = log | x-y | +C " "` ...(i)
It is given when x = 0 , =-1
` :. 0+(-1)= log (0+1)+C rArr C = -1 `
On substituting this value in Eq. (i) we get required prticular solution as ` x+y = log | x - y | -1 `
` rArr loh | x-y| = x+ y +1 `
Discussion
76.

The differential equation of all straight lines passing through the point (1,-1) isA. `y=(x+1)(dy)/(dx)+1`B. `y=(x+1)(dy)/(dx)-1`C. `y=(x-1)(dy)/(dx)+1`D. `y=(x-1)(dy)/(dx)-1`

Answer» Correct Answer - D
Discussion
77.

Find the particular solution of the differential equation `(1+e^(2x))dy+(1+y^2)e^x dx=0,`given that `y=1`when `x=0.`A. `tan^(-1)y+tan^(-1)e^(x)=pi/2`B. ` tan^(-1)x+tan^(-1)e^(y)=pi/2`C. ` tan^(-1) x + tan^(-1)e^(y)=pi/4`D. ` tan^(-1)y +tan^(-1)e^(x)=pi/3`

Answer» Correct Answer - a
Given differential equation is
` 1+e^((2x)) dy + ( 1+y^(2)) e^(x)dx = 0 `
Separating the variables , we get ` (dy)/(1+y^(2)) +(e^(x)dx)/(1+e^(2x))=0`
On intergrating both sides , we get ` int (dy)/(1+y^(2)) + int (e^(x)dx)/(1+e^(2x)) = C`
Put `e^(x) = t rArr e^(x) dx = dt `
` rArr tan^(-1) y + int (dt)/(1+t^(2))=C`
` rAr tan^(-1) ty + tan^(-1) t = C`
` rArr tan^(-1) + tan^(-1) e^(x) = C " "` ... (i) ltbRgt Now, put x = 0 and y = 1
` :. tan^(-1) + tan^(-1) e^(0) = C`
` rArr pi/4 +pi/4 C rArr C = pi/2 `
Discussion
78.

The solution of ` (dy)/(dx) = 1+ y + y^(2) + x+ xy + xy^(2)` isA. `tan^(-1)((2y+1)/(sqrt(3)))=x+x^(2)+C`B. ` 4 tan^(-1) ((4y+1)/(sqrt(3)))= sqrt(3)(2x+x^(2))+C`C. ` sqrt(3) tan^(-1) ((3y+1)/3)=4 (1+x+x^(2))+C`D. ` 4 tan^(-1) ((2y+1)/(sqrt(3)))= sqrt(3)(2x+x^(2))+C`

Answer» Correct Answer - a
Given, ` (dy)/(dx) = (1+y+y^(2))+ x(1+y+y^(2))`
` = ( 1+ y + y^(2)) (1+x) `
` rArr = (dy)/(1+y+y^(2)) = (1+x)`
` rArr int (dy)/((y+1/2)^(2)+((sqrt3)/2)^(2)) = int (1+x)dx`
` = (1+y+y^(2)) (1+x)`
` rArr (dy)/(1+y+y^(2)) = (1+x)dx`
` rArr int (dy)/((y+1/2)^(2)+((sqrt(3))/2)^(2) )= int (1+x)dx`
` rArr 1/(sqrt(3)/2) tan^(-1) ((2y+1)/(sqrt(3))) = sqrt(3) (2x +x^(2)) +C`
Discussion
79.

The solution of the differential equation `y dx - (x + 2y^2)dy=0` is `x=f(y)`. If `f(-1)=1,` then `f(1)` is equal toA. 4B. 3C. 1D. 2

Answer» Correct Answer - b
Given, `y dx = (x - 2y^(2)) dy = 0 `
` rArr ( y dx - x dx)/(y^(2)) = 2dy rArr d ( x/y) = 2dy `
On integrating both sides , we get
` rArr int d(x/y) = 2 int dy `
` rArr x/y = 2y +C`
At x = 1 , y = -1
` :. C = 1 `
Now, ` x/y = 2y +1`
When y = 1 , then f(1) = 3
Discussion
80.

Find the general solution of the differential equations `(e^x+e^(-x))dy-(e^x-e^(-x))dx=0`A. ` y = log |e^(x)=e^(-x)|+C`B. ` y = log. |(e^(x)-e^(-x))/(e^(x)+e^(-x))|+C`C. `y= log|e^(x)+e^(-x)|+C`D. None of these

Answer» Correct Answer - b
Given ` (e^(x)+e^(-x))dy - (e^(x)-e^(-x))dx = 0 `
` rArr (e^(x) +e^(-x))dy = (e^(x)-e^(-x))dx`
On seperating the variables , we get ` dy = ((e^(x)-e^(-x))/(e^(x)+e^(-x)))dx`
On integrating , we get ` int dy = int ((e^(x)-e^(-x))/(e^(x)+e^(-x)))dx`
Let `e^(x) +e^(-x) = t rArr (e^(x) -e^(-x)) dx =dt`
` :. int dy = int (e^(x) - e^(-x))/t (dt)/(e^(x)-e^(-x))=int 1/t dt`
` rArr y = log |t| +C`
` rArr y = |log| |e^(x) +e^(-x) | +C`
Discussion
81.

Findthe equation of the curve passing through the point `(0,pi/4)`whose differential equation is `s in" "x" "cos" "y" "dx" "+" "cos" "x" "s in" "y" "dy" "=" "0`.A. ` cos y = (cos a)/(sqrt(2))`B. ` cos y = (sin x)/2`C. ` cos y = (sec x)/(sqrt(2))`D. ` cos y = ("cosec"x)/(sqrt(2))`

Answer» Correct Answer - c
The differential equation of the given curve is
`sin x cos y dx +cos x sin y dy =0`
`rArr (sinx)/(cosx) dx +(siny)/(cosy) dy = 0`
`rArr tan x dx + tan y dy = 0 `
On integrating both sides , we get
` int tan x dx _ int tan y dy = log C`
` rArr log ( sec x) + log (sec y ) = log C`
` sec x sec y = C " " ` ...(i)
The curve passes through the point `(0,pi/4)` ,therefore put
` x = 0 , y = pi/4,` we get ` sec 0 sec pi/4 = C`
` rArr C = sqrt(2)`
On putting the value of C in Eq. (i) , we get
` sec x * sec y = sqrt(2)`
` rarr sec x* 1/(cosy) = sqrt(2)`
` rArr cos y = (sec x)/(sqrt(2))`
Hence, the required equation of the curve is
` cos y =(secx)/(sqrt(2))`
Discussion
82.

The differential eqution for `(x)/(a)+(y)/(b)=1` isA. `(dy)/(dx)=0`B. `(d^(2)y)/(dx^(2))=0`C. `(d^(2)y)/(dx^(2))+(dy)/(dx)=0`D. `(d^(2)y)/(dx^(2))-(dy)/(dx)=0`

Answer» Correct Answer - B
Discussion
83.

The equation of the curve satisfying the eqution `(xy-x^(2)) (dy)/(dx) = y^(2)` and passing through the point `(-1,1)` isA. `y=(logy-1)x`B. `y=(log y+1)x`C. `x= (log x-1)y`D. `x = (log x+1)y`

Answer» Correct Answer - b
We have , `(xy-x^(2)) (dy)/(dx) = y^(2)`
` rArr y^(2) (dx)/(dy) = xy -x^(2)`
` rArr 1/(x^(2)) (dx)/(dy) - 1/x . 1/y = -1/(y^(2))`
Put `1/x = v`
` rArr -1/(x^(2)) (dx)/(dy) = (dv)/(dy)`
` (dv)/(dy) + v/y = 1/(y^(2))`, which is linear
` :. IF - e^(int1/ydy) = e^(logy) = y`
` :. " The solution " vy = int 1/(y^(2) ) . y dy + C`
` rArr y/x = log y + C rArr y = x ( log y + C)`
This passes through the point `(-1),1)`
` :. 1 = -1 ( log 1 +C) " i.e " C =-1`
Thus , the equation of the curve is y = x `(log y -1)`
Discussion
84.

The tangent at any point `(x , y)`of a curve makes an angle `tan^(-1)(2x+3y)`with x-axis. Find the equation of the curve if itpasses through (1,2).A. ` 6x+9y +2 = 26 e^(3(x-1))`B. ` 6x- 9y +2 = 26 e^(3(x-1))`C. ` 6x + 9y -2 = 26 e^(3(x-1))`D. ` 6x - 9y -2 = 26 e^(3(x-1))`

Answer» Correct Answer - a
Given , `(dy)/(dx) = tan theta = 2x +3y`
Put ` 2x + 3y = z rArr 2 +3 (dy)/(dx) = (dz)/(dx)`
` rArr (dy)/(dx) = ((dz)/(dx) - 2 ) 1/3 `
` :. (dz)/(dx) - 2 = 3z rArr (dz)/(3z + 2) = dx`
On integrating , we get
`(log (3z+2))/3 = x+C`
` rArr (log (6x + 9y +2))/3 = x=C`
Since , it passes through (1,2)
` :. (log (6+18+2))/3 = 1+C rArr C = (log 26)/3 -1`
` :. (log (6x +9y+2))/3 = x + (log 26)/3 -1`
` rArr log ((6x+9y+2)/26) = 3 ( x-1)`
` rArr 6x + 9y +2 = 26 e^(3(x-1))`
Discussion
85.

The solution of the differential eqution `(d^(2)y)/(dx^(2)) = e^(-2x)` is ` y = c_(1)e^(-2x) +c_(2)x + c_(3)` where `c_(1)` isA. 1B. `1/4`C. `1/2`D. 2

Answer» Correct Answer - b
Given , ` (d^(2)y)/(dx^(2)) = e^(-2x)`
` rArr (dy)/(dx) = (e^(-2x))/(2) +c_(2)` [ integrating ]
`rArr y = (e^(-2x))/4 + c_(2)x + c_(3)` [ integrating]
But ` y = c(1)e^(-2x) + c_(2) x + c_(3) ` [given ]
` :. C_(1) = 1/4 `
Discussion
86.

Let y = f(x) be a curve passing through `(e,e^( e ))` which satisfy the differential equation ` ( 2ny + xy log_(e) x ) dx - x log_(e) x dy = 0 , x gt 0, y gt 0`A. eB. 1C. 0D. 2

Answer» Correct Answer - c
`2ny + xy log _(e)x) dx = x log_(e) x dy`
` rArr (dy)/y = ((2n)/(x log _(e)x)+1) dx`
` rArr log (y) = 2n log | log x| + x+C`
` :. " Curve through" (e,e^(e )) ` so , C = 0
` :. y = e^(x+log(logx)2n)`
` rArr f(x) = e^(x) (log x )^(2n) `
Now, `g(x) =underset(nrarroo)lim f(x)={{:(oo","," if",xlt(1)/(e)),(0","," if ",(1)/(e)lt xlt e),(oo","," if ",x gt e):}`
`therefore" "int_(1//e)^(e)g(x)dx=0`
Discussion
87.

The solution of the differential equation `xy(dy)/(dx)={(1+y^2)(1+x+x^2)}/(1+x^2)` is:A. ` sqrt(1+y^(2))=Cxe^(tan^(-1)x)`B. ` sqrt(1-y^(2))=Cxe^(tan^(-1)x)`C. ` sqrt(1+y^(2))=Ce^(tan^(-1)x)`D. ` sqrt(1+y^(2))=Cx`

Answer» Correct Answer - a
` xy (dy)/(dx) = (1 +y^(2)) (1+x/(1+x^(2)))`
` rArr y/(1+y^(2)) (dy)/(dx) = 1/x + 1/(1+x^(2))`
` 1/2 In (1+y^(2))= In x + tan^(-1) x In C`
` rArr sqrt(1+y^(2)) = Cxe ^(tan^(-1)x)`
Discussion
88.

The equation of curve through point (1,0) which satisfies the differential equation `(1+y^(2))dx- xy ` dy = 0 , isA. `x^(2)+y^(2)=4`B. `x^(2)-y^(2)=1`C. `2x^(2)+y^(2)=2`D. None of the above

Answer» Correct Answer - c
Given , `(dx)/(x) = (ydx)/(1+y^(2))`
` rArr log x = 1/2 log (1+y^(2))+log C rArr x = C sqrt(1+y^(2))`
But it passes through (1,0) ,so we get C = 1
` :. " Solution is " x^(2) - y^(2) = 1`
Discussion
89.

The solution of the differential equation `(dy)/(dx) + y = 1 ( y != x)` isA. `y=ce^(x)`B. `y = ce^(-x)`C. `y = 1+ce^(x)`D. `y = 1+ce^(-x)`

Answer» Correct Answer - a
The given equation is `(dy)/(dx) +y =1`
` rArr (dy)/(dx) = 1-y rArr (dy)/(1-y) = dx`
On integrating both sides , we get
` int (dy)/(1-y) = int dx`
` rArr log.((1-y))/-1 = x+c_(1)` ,
` rArr (1-y) = e^(-xc_(1))`
`rArr 1-y = e^(-x) .e^(-c_(1))`
`rArr y = 1 +ce^(-x), "where" c = e^(-c_(1))`
Discussion
90.

Find the general solution of each of the following differential equations: `(dy)/(dx)=(1-cos x)/(1+cosx)`A. ` y = tan.x/2+c`B. `y = 2 tan.x/2 +c`C. ` y = 2 tan. x/2 -x +c`D. ` y = 2tan. x/2 +x + c`

Answer» Correct Answer - c
The given equation is `(dy)/(dx) = (1-cosx)/(1+cosx)`
` rArr (dy)/(dx) = (2 sin^(2) x//2)/(2 cos^(2)x//2)`
` rArr (dy)/(dx) = tan^(2) x//2`
` rArr dy = (tan^(2)x//2) dx`
On integratng both sides , we get
` int dy = int tan^(2) x//2 dx`
` = int (sec^(2). x/2 -1)dx`
` = (tan.x/2)/(1//2)-x +c`
` rArr y = 2 tan. x/2 - x +c`
Discussion
91.

`x +(dy)/(dx) = sqrt(1+ ((dy)/(dx))^(2))`A. 1 and 2B. 1 and 1C. `1 and (1)/(2)`D. 2 and 1

Answer» Correct Answer - B
Discussion
92.

The solution of the differential equation `(dx)/x +(dy)/y = 0` isA. xy=CB. `x+y=C`C. `log x log y=C`D. `x^(2)+y^(2)=C`

Answer» Correct Answer - a
The given differential equation is
`(dx)/x+(dy)/y =0`
On integrating both sides , we get
` log x + log y = k rArr log xy = k` ltbRgt ` rArr xy =e^(k) rArr xy =C,`
where `C = e^(k)`
Discussion
93.

Order and degree of the differential equation `y(dy)/(dx)=(x)/((dy)/(dx) + ((dy)/(dx))^(3))` respectively areA. 1 and 1B. 1 and 2C. 1 and 3D. 1 and 4

Answer» Correct Answer - D
Discussion
94.

Order and degree of the differential equation `(dy)/(dx)=(2sinx+3)/((dy)/(dx))` respectively areA. 1 and 1B. 1 and 2C. 2 and 1D. 2 and 2

Answer» Correct Answer - B
Discussion
95.

Solution of differential equation `(dy)/(dx)=xlogx` isA. `y=x^(2)log|x|-(x^(2))/(2)+c`B. `y=(x^(2))/(2)log|x|-x^(2)+c`C. `y=(x^(2))/(2)+(x^(2))/(2)log|x|+c`D. `y=(x^(2))/(2)log|x|-(x^(2))/(4)+c`

Answer» Correct Answer - D
Discussion
96.

The solution of differential equation `(1+x) y dx +(1-y) x dy = 0 ` isA. ` log_(e)(xy)+x-y=C`B. ` log _(e)(x/y)+x+y = C`C. ` log_(e) (x/y)-x+y = C`D. `log_(e) (xy)-x+y = C`

Answer» Correct Answer - a
Given , `(1+x) y dx +(1-y) x dy = 0 `
` rArr ((1-y))/y dy +((1+x))/x dx = 0 `
On integrating both sides , we get
` rArr int (1/y -1) dy +int (1/x +1) dx = 0 `
` rArr log_(e)y - y + log_(e) x + x =C`
` rArr log_(e) (xy) +x-y = C`
Discussion
97.

The general solution of the differential equation ` (dy)/(dx) = ((1+y^(2)))/(xy(1+x^(2)))` isA. `(1+x^(2))(1+y^(2))=C`B. `(1+x^(2))(1+y^(2))=Cx^(2)`C. `(1-x^(2))(1-y^(2))=C`D. `(1+x^(2))(1+y^(2))=Cy^(2)`

Answer» Correct Answer - b
Given differential equation can be rewritten as
` y/(1+y^(2))dy= (dx)/(x(1+x^(2)))`
`rArr 1/2 int (2y)/(1+y^(2))dy = (dx)/(x(1+x^(2)))`
` rArr 1/2 int (2y)/((1+y^(2)))dy=1/2 int (dt)/(t(t+1))`
[ put `x^(2)` = t and 2x dx = dt in RHS integral ]
` rArr 1/2 int ((2y" "dy))/(1+y^(2))=1/2 int (1/t - 1/(1+t))dt`
` rArr 1/2 * log (1+y^(2)) = 1/2 [ log t - log (1+t)] +1/2 log C`
` rArr log 1+y^(2) = logx^(2)- log (1+x^(2))+log C`
` rArr log (1+y^(2))(1+x^(2))= log Cx^(2)`
` rArr (1+y^(2))(1+x^(2))= Cx^(2)`
Discussion
98.

The solution of the differential equation ` (dy)/(dx)= x log x ` isA. ` y=x^(2)logx-(x^(2))/2 +C`B. `y=(x^(2))/x log x-(x^(2))/4+C`C. `y = (x^(2))/2+(x^(2))/2 log x +C`D. None of these

Answer» Correct Answer - c
Given `dy = x log x dx`
On integrating both sides , we get
` rArr y = (x^(2))/2 log x - int x/2 dx `
` rArr y = (x^(2))/2 logx - (x^(2))/4 +C`
Discussion
99.

The solution of `(dy)/(dx) + sqrt(((1-y^(2))/(1-x^(2))))=0`isA. `tan^(-1)x +cot^(-1)x = C `B. `sin^(-1)x + sin^(-1)y=C`C. `sec^(-1)x+"cosec"^(-1)x=C`D. None of these

Answer» Correct Answer - b
Given ` dy/dx + sqrt((1-y^(2))/(1-x^(2))) = 0 `
` rArr int (dy)/(sqrt(1-y^(2)))+ int (dx)/(sqrt(1-x^(2)) )=0`
` rArr sin^(-1) y + sin^(-1) x = C`
Discussion
100.

The solution of differential edquation `(dy)/(dx) +1 = cosec (x+y)` isA. `cos(x+y)+x=C`B. `cos(x+y)=C`C. `sin(x+y)+x=C`D. `sin(x+y)+sin(x+y)=C`

Answer» Correct Answer - b
Given `(dy)/(dx) +1 = " cosec" (x+y)`
Put ` x+y = t`
` rArr 1+ (dy)/(dx) = (dt)/(dx)`
` :. (dt)/("cosec t ") = dx `
` rArr int sin t dt = int dx `
` rArr cos ( x+y) +x = C`
Discussion