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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.
| 101. |
The differential equation representing all possible curves that cut each member of the family of circles `x^(2)+y^(2)-2Cx=0` (C is a parameter) at right angle, isA. `(dy)/(dx)=(2xy)/(x^(2)+y^(2))`B. `(dy)/(dx)=(2xy)/(x^(2)-y^(2))`C. `(dy)/(dx)=(x^(2)+y^(2))/(2xy)`D. `(dy)/(dx)=(x^(2)-y^(2))/(2xy)` |
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Answer» Correct Answer - B Here, we have to find the orthogonal trajectories of the family of circles `x^(2)+y^(2)-2Cx=0" …(i)"` Differentiating (i) w.r.t. x, we get `2x+2y(dy)/(dx)-2C=0rArrC=x+y(dy)/(dx)" ...(ii)"` From (i) and (ii), we obtain `x^(2)+y^(2)-2x(x+y(dy)/(dx))=0" [By eliminating C]"` `rArr" "y^(2)-x^(2)-2xy(dy)/(dx)=0rArry^(2)-x^(2)=2xy(dy)/(dx)" ...(iii)"` This is the differential equation representing the given family of circles. To find the differential equation of the orthogonal trajectories, we replace `(dy)/(dx)by-(dx)/(dy)` in equation (iii). Thus, the differential equation representing the orthogonal trajectories is `y^(2)-x^(2)=-2xy(dx)/(dy)or, (dy)/(dx)=-(2xy)/(x^(2)-y^(2))` |
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| 102. |
The population of a village increases continuouslyat the rate proportional to the number of its inhabitants present at anytime. If the population of the village was 20, 000 in 1999 and 25000 in theyear 2004, what will be the population of the |
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Answer» Let the population at time `t` be `y`, then `(dy)/(dt)prop y` `implies (dy)/(dt)=ky`, where `k` is a constant `implies (dy)/(y)=k dt` On integration, `logy=kt+C`……..`(1)` In year `1999`, `t=0`, `y=20000` from equation `(1)`, `log20000=k(0)+C` `implies log20000=C`.......`(2)` In year `2004`, `t=5`, `y=25000` Therefore, from equation `(1)`, `log25000=k*5+C` `implies log25000=5k+log20000` [from equation `(2)` ] `implies 5k=log((25000)/(20000))=log((5)/(4))` `implies k=(1)/(5)log(5)/(4)` For the year `2009`, `t=10`years Now, put the values of `t`, `k` and `C` in equation `(1)`, `logy=10xx(1)/(5)log((5)/(4))+log(20000)` `implies logy=log[20000xx(5)/(4))^(2)]` `implies y=20000xx(5)/(4)xx(5)/(4)impliesy=31250` Therefore, the population of the village in the year `2009` will be `31250`. |
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| 103. |
In aculture, the bacteria count is 1,00,000. The number is increased by 10% in 2 hours.In how many hours will the count reach 2,00,000, if the rate of growth of bacteriais proportional to the number present? |
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Answer» Let at time `t`, the number of bacteria be `y` then `(dy)/(dt)propyimplies (dy)/(dt)=ky`, where `k` is proportional constant. `implies (dy)/(y)=k dt` `impliesint(dy)/(y)=intk dt` `implies log|y|=kt+C` .........`(1)` Initially at `t=0`, `y=100000` `:. log100000=C` ........`(2)` when `t=2`, `y=110000` `log110000=2k+C` ..........`(3)` `:.` Subtracting equation `(2)` from equation `(3)`, `log110000-log100000=2k` `implies log"(110000)/(100000)=2k` `implies k=(1)/(2)log((11)/(10))` put the value of `k` and `C` in equation `(1)`, `logy=(1)/(2)log((11)/(10))t+log100000` when `y=200000` then `log200000=(1)/(2)((11)/(10))t+log100000` `implies log((200000)/(100000))=(1)/(2)log((11)/(10))t` `implies 2log(2)=log((11)/(10))t` `implies t=(2log2)/(log((11)/(10)))` Therefore, the number of bacteria will increase from `100000` to `200000` in `(2log2)/(log((11)/(10)))` time. |
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| 104. |
A spherical balloon is filled with 4500p cubic meters ofhelium gas. If a leak in the balloon causes the gas to escape at the rate of `72pi`cubic meters per minute, then the rate (in metersper minute) at which the radius of the balloon decreases 49 minutes after theleakage began is(1) `9/7`(2) `7/9`(3) `2/9`(4) `9/2`A. `6//7`B. `4//9`C. `2//9`D. None of these |
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Answer» Correct Answer - C `V=4/3pir^(3)` Initially volume of gas in balloon is 4500 `pi` cubic meters. `therefore 4500 pi=(4pir^(3))/(3)` `therefore r=15` m. Given `(dV)/(dt) = 72pi` cubic meter per minute `therefore` After 49 minutes, volume of gas leaked `=72pi xx 49` `therefore` Volume of gas in balloon after 49 minutes `=(4500-49xx72)pi=972pim^(3)` `therefore 972pi=(4pir^(3))/(3)` `therefore r^(3)=3 xx 243 = 3 xx 3^(5)` `therefore r=9` Now, `(dV)/(dt) = 4pir^(2) (dr)/(dt)` For the rate of decrease in radius after 49 minutes. `72pi = 4pi xx 9 xx 9(dr)/(dt)` `(dr)/(dt) 2/9` meter per minute |
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| 105. |
The population p(t) at time t of a certain mouse species satisfies thedifferential equation `(d p(t)/(dt)=0. 5 p(t)-450`If `p(0)""=""850`, then the time at which the population becomes zero is(1) 2 ln 18 (2) ln9 (3) `1/2`In 18 (4) ln 18A. 2 ln 18B. ln 9C. `1/2` ln 18D. ln 18 |
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Answer» Correct Answer - A We have, `2(dp(t))/(900-p(t))=-dt` Integrating, we get `-2"ln "(900-p(t))=-t+c` when `t=0, p(0)=850` `therefore -2" ln "(50/(900-p(t)))=-t` `therefore 900-p(t)=50e^(t//2)` Let `p(t_(1))=0` `therefore 0=900-50e^(t//2)` `therefore t_(1)=2` ln 18 |
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| 106. |
Which of the following statements on ordinary differential equations is/are true ? (i) The number of arbitrary constants is same as the degree of the differential equation. (ii) A linear differential equation can contain products of the dependent variable and its derivatives. (iii) A particular integral cannot contains arbitrary constants. (iv) By putting `v=(y)/(x)` any homogeneous first order differential equation transforms to variable separable form.A. (i) and (iii) onlyB. (ii) and (iii) onlyC. (iii) onlyD. (i) and (iv) only |
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Answer» Correct Answer - C The number of arbitrary constants in the solution of a differential equation is same as its order. So, statements (i) is not true. Statement (ii) is not true. However, statement (iii) is true. Some homogeneous differential equations reduce to variable separable form by putting `v=(y)/(x)`. |
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| 107. |
The differential equation `(dy)/(dx)+(y^(2))/(x^(2))=(y)/(x)` has the solutionA. `x=y(logx+C)`B. `y=x(logy+C)`C. `x=(y+C)logx`D. `y=(x+C)logy` |
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Answer» Correct Answer - A We have, `(dy)/(dx)+(y^(2))/(x^(2))=(y)/(x)` Putting `y=vx` and `(dy)/(dx)=v+x(dv)/(dx)`, we get `v+x(dv)/(dx)+v^(2)=v rArr x(dv)/(dx)=-v^(2)rArr-(1)/(v^(2))dv=(1)/(x)dx` On integrating, we get `(1)/(v)=log x+C rArr x=y(logx+C)` |
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| 108. |
On putting `(y)/(x)=v` the differential equation `(dy)/(dx)(2xy-y^(2))/(2xy-x^(2))` is transfored toA. `x(2v-1)dx=3v(v-1)dx`B. `x(2v-1)dv=3v(1-v)dx`C. `x(1-2v)dv=(v^(2)-2v)dx`D. `(1-2v)dv=(v^(2)-2v)dx` |
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Answer» Correct Answer - B Putting `y=vx` and `(dy)/(dx)=v+(dv)/(dx)`, given differetial equation reduces to `v+x(dv)/(dx)=(2v-v^(2))/(2v-1)rArr3v(1-v)dx=x(2v-1)dv` |
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| 109. |
The differential equation of all non-horizontallines in a plane is(a)`( b ) (c) (d)(( e ) (f) d^(( g )2( h ))( i ) y)/( j )(( k ) d (l) x^(( m )2( n ))( o ))( p ) (q) (r)`(s)(b) `( t ) (u) (v)(( w ) (x) d^(( y )2( z ))( a a ) x)/( b b )(( c c ) d (dd) y^(( e e )2( f f ))( g g ))( h h ) (ii)=0( j j )`(kk)(c)`( d ) (e) (f)(( g ) dy)/( h )(( i ) dx)( j ) (k)=0( l )`(m) (d) `( n ) (o) (p)(( q ) dx)/( r )(( s ) dy)( t ) (u)=0( v )`(w)A. `(d^(2)y)/(dx^(2))`B. `(d^(2)x)/(dy^(2))=0`C. `(dy)/(dx)=0`D. `(dx)/(dy)=0` |
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Answer» Correct Answer - B The general equation of all non-horizontal lines in xy-plane is `ax+by=1`, where `a ne 0`. Now, `ax+by=1` `rArr" "a(dx)/(dy)+b=0" [Diff. w.r. to y]"` `rArr" "a(d^(2)x)/(dy^(2))=0" [Diff. w.r. to y]"` `rArr" "(d^(2)x)/(dy^(2))=0" "[because a ne 0]` Hence, the required differential equation is `(d^(2)x)/(dy^(2))=0` |
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| 110. |
Solve the following differential equation:`((x^2-1)dy)/(dx)+2x y=1/(x^2-1); |x| !=1` |
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Answer» Given different equation is `(x^(2)-1)(dy)/(dx)+2xy=(1)/(x^(2)-1)` `Rightarrow (dy)/(dx)+((2x)/(x^(2)-1))y=(1)/((x^(2)-1)^(2))` Which is linear differential equation. On comparing it with `(dy)/(dx)+Py=Q`, we get `P=(2x)/(x^(2)-1).Q=(1)/((x^(2)-1)^(2))` `IF=e^(intpdx)=e^(int((2x)/(x^(2)-1))^(dx))` `"Put" x^(2)-1=t Rightarrow 2xdx=dt` `therefore IF=r^(int(dt)/(t))=e^(log t)=t=(x^(2)-1)` The complete solution is `y.IF=int Q.IF+K` `Rightarrow y.(x^(2)-1)=int(1)/((x^(2)-1)^(2)).(x^(2)-1)dx+K` `Rightarrow y.(x^(2)-1)=int(dx)/((x^(2)-1))+K` `Rightarrow y.(x^(2)-1)=(1)/(2)log ((x-1)/(x+1))+K` |
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| 111. |
Find the solution of `(dy)/(dx)=2^(y-x)` |
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Answer» `"Given that",=(dy)/(dx)=2^(y-x)` `Rightarrow (dy)/(dx)=(2^(y))/(2^(x))[therefore 2^(m-n)=(a^(m))/(a^(n))]` `Rightarrow (dy)/(2^(y))=(dx)/(2^(x))` On integrating both sides, we get `int2^(-y)dy=int2^(-x)dx` `Rightarrow (-2^(-y))/(log2)=(-2^(-x))/(log2)+C` `Rightarrow -2^(-y)+2^(-x)=+Clog2` `Rightarrow 2^(-x)+2^(-y)=-Clog2` `Rightarrow 2^(-x)+2^(-y)=K` |
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| 112. |
Show that `y=a*e^(2x)+b*e^(-x)` is a solution of the differential equation `(d^(2)y)/(dx^(2))-(dy)/(dx)-2y=0`. |
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Answer» `y=ae^(2x)+be^(-x)`……….`(1)` Differentiate with respect to `x` `(dy)/(dx)=a*e^(2x)*(2)+b*e^(-x)(-1)` `implies (dy)/(dx)=2a*e^(2x)-b*e^(-x)`………`(2)` Again differentiate with respect to `x` `(d^(2)y)/(dx^(2))=2a*e^(2x)*(2)-b*e^(-x)(-1)` `implies (d^(2)y)/(dx^(2))=4a*e^(2x)+b*e^(-x)`..............`(3)` Now `L.H.S=(d^(2)y)/(dx^(2))-(dy)/(dx)-2y` `=(4a*e^(2x)+be^(-x))` `-(2ae^(2x)-b*e^(-x))-2(ae^(2x)+be^(-x))` [Putting the values from `(1)`, `(2)` and `(3)`] `=4ae^(2x)+be^(-x)-2ae^(2x)+be^(-x)-2ae^(2x)-2be^(-x)` `=0=R.H.S`. `:. y=ae^(2x)+be^(-x)` is a solution of the given differential equation. |
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| 113. |
If `dy/dx=e^(-2y) and y=0` when `x=5,` then the value of x for `y=3` is |
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Answer» Given that, `(dy)/(dx)=e^(-2 y)Rightarrow (dy)/(e^(-2y))=dx` `Rightarrow inte^(2y)=intdx Rightarrow (e^(2y))/(2)=x+C ....(i)` When x=5 and y=0, then substituting these values in Eq. (i), we get `(e^(0))/(2)=5+C` `Rightarrow (1)/(2)=5+C Rightarrow (1)/(2)-5=(9)/(2)` `"Eq. = (i) becomes e"^(2y)=2x-9` When, y=3, then `e^(6)=2x-9 Rightarrow 2x=e^(6)+9` `x=((e^(6))+9)/(2)` |
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| 114. |
Prove that `y=e^(x)+m` is a solution of the differential equation `(d^(2)y)/(dx^2)-(dy)/(dx)=0`, where `m` is a constant. |
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Answer» `y=e^(x)+m`……..`(1)` Differentiate with respect to `x` `(dy)/(dx)=e^(x)`……….`(2)` Again differentiate with respect to `x` `(d^(2)y)/(dx^(2))=e^(x)` `implies (d^(2)y)/(dx^(2))=(dy)/(dx)` [From eq. `(2)`] `implies (d^(2)y)/(dx^(2))-(dy)/(dx)=0` `:. y=e^(x)+m` is a solution of the given differential equation. |
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| 115. |
Find the differential equation of all non-vertical lines in a plane. |
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Answer» Since, the mainly of all non vertical line is y=mx+c, where `mne "tan"(pi)/(2)` On differentiating w.r.t.x. , we get `(dy)/(dx)=m` Again, differentiating w.r.t. x, we get `(d^(2)y)/(dx^(2))=0` |
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| 116. |
The differential equation of all non-vertical lines in a plane, isA. `(d^(2)y)/(dx^(2))=0`B. `(d^(2)x)/(dy^(2))=0`C. `(dy)/(dx)=0`D. `(dx)/(dy)=0` |
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Answer» Correct Answer - A The general equation all non-vertical lines in a plane is `ax+by=1`, where `b ne 0`. Now, `ax+by=1` `rArr" "a+b(dy)/(dx)=0" [Differentiating w.r. to x]"` `rArr" "b(d^(2)y)/(dx)=0" [Differentiating w.r. to x]"` `rArr" "(d^(2)y)/(dx^(2))=0" "[because b ne 0]` Hence, the differential equation is `(d^(2)y)/(dx^(2))=0` |
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| 117. |
Show that `y=x sin x` is a solution of the differential equation `(d^(2)y)/(Dx^(2))+y-2cosx=0`. |
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Answer» `y=x sin x`……….`(1)` Differentiate with respect to `x` `(dy)/(dx)=x*cosx+sinx`………..`(2)` Again differentiate with respect to `x` …………`(2)` Again differentiate with respect to `x` `(d^(2)y)/(dx^(2))=x(-sinx)+cosx+cosx` `=-x sin x+2cosx` `=-y +2cos x` [From eq. `(1)`] `implies (d^(2)y)/(dx^(2))+y-2cosx=0` `:. y=sinx` is a solution of the given differential equation. |
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| 118. |
Find the order and degree (if defined) of the equation:`a=(1[1+((dy)/(dx))^2]^(3/2))/((d^2y)/(dx^2)),`where `a`is constant |
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Answer» Correct Answer - Order-2, degree-2 The given equation when exposed as a polynomial in derivative as `a^(2)(d^(2)y)/(dx^(2))^(2)=[1+((dy)/(dx))^(2)]^(3)` Clearly, it is second-order differential equation of degree 2. |
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| 119. |
Find the differential equation ofAll-horizontal lines in a plane.All non-vertical lines in a plane. |
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Answer» Correct Answer - `(d^(2)y)/(dx^(2))=0` Alll such lines are given by `y=mx+c`. Here, we have two effective constants m and c. So, it is required to differentiate twice. `y=mx+c` `therefore (dy)/(dx)=m` or `(d^(2)y)/(dx^(2))=0` |
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| 120. |
`((d^(4)y)/(dx^(4)))^(3)+3((d^(2)y)/(dx^(2)))^(6) + sinx =2cosx` |
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Answer» Correct Answer - Order-4, degree-3 `((d^(4)y)/(dx^(4)))^(3)+3((d^(3)y)/(dx^(2)))^(6)+sinx=2cosx` Clearly, order is 4 and degree is 3. |
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| 121. |
Find the order and degree (if defined) of the equation:`((d^3y)/(dx^3))^(2/3)+4-3(d^2)/(dx^2)+5(dy)/(dx)=0` |
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Answer» Correct Answer - Order-3, degree-2 We have `(d^(3)y)/(dx^(3))^(2//3)+4-3(d^(2)y)/(dx^(2))+5(dy)/(dx)=0` or `(d^(3)y)/(dx^(3))^(2)=(3(d^(2)y)/(dx^(2))-5(dy)/(dx)-4)^(3)` Clearly, it is differential equation of degree 2 and order 3. |
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| 122. |
Find the order and degree (if defined) of the equation:`(d^4y)/(dx^4)-sin((d^3y)/(dx^3))=0` |
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Answer» Correct Answer - Order-4, degree not defined `(d^(4)y)/(dx^(4))-sin(d^(3)y)/(dx^(3))=0` The highest-order derivative present in the differential equation is `(d^(4)y)/(dx^(4))`. Thus, its order is four. However, the given diifferential equation is not a polynomial equation, Hence, its degree is not defined. |
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| 123. |
The order and degree of differential equation of all tangent lines to parabola `x^2=4y` isA. 1,2B. 2,2C. 3,1D. 4,1 |
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Answer» Correct Answer - A The equaiton of any tangent to `x^(2)=4y` is `x=my+(1)/(m)`, where m is an arbitrary constant. Differentiating this w.r.t. to x, we get `1=m(dy)/(dx)rArr m=(1)/((dy)/(dx))` Putting the value of m in x = `my+(1)/(m)`, we get `x=(y)/((dy)/(dx))+(dy)/(dx)rArr((dy)/(dx))^(2)-x(dy)/(Dx)+y=0`, Which is differential equation of order 1 and degree 2. |
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| 124. |
Find the order and degree (if defined) of the equation:`(d^3y)/(dx^3)=x ln((dy)/(dx))` |
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Answer» Correct Answer - Order-3, degree-2 `(d^(3)y)/(dx^(3))=x" ln "(dy)/(dx)` Clearly, the order is 3 and the degree is not defined due to the `" ln "(dy)/(dx)` term |
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| 125. |
The degree of the differential equation `x=1+((dy)/(dx))+1/(2!)((dy)/(dx))^2+1/(3!)((dy)/(dx))^3+.............` (A) 3 (B) 2 (C) 1 (D) not definedA. 3B. 1C. not definedD. none of these |
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Answer» Correct Answer - B We have, `x=e^((dy)/(dx))rArrlogx=(dy)/(dx)` Clearly, it is differential of degree one. |
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| 126. |
Form the differential equation of family of lines concurrent at the origin. |
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Answer» Equation of family of line concurrent at origin is given by y=mx `therefore (dy)/(dx)=m` Putting the value of m in (1), we get `y=(dy)/(dx).x` or `xdy-ydx=0`, which is required differential equation of order q. Here order is same as number of effective constants in the equation of family of curve. |
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| 127. |
Find the order and degree (if defined) of the following differential equations: `(d^(2)y)/(dx^(2)) = {1+((dy)/(dx))^(4)}^(5//3)` |
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Answer» Correct Answer - Order-2, degree-3 `(d^(2)y)/(dx^(2))={1+((dy)/(dx))^(4)}^(5//3)` or `((d^(2)y)/(dx^(2)))^(3)={1+((dy)/(dx))^(4)}^(5)` Hence, the order is 2 and degree is 3. |
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| 128. |
Find the order and degree of the following differential equations. i) `(dy)/(dx)+y=1/((dy)/(dx))`, ii) `e^(e^(3)y)/(dx^(3))-x(d^(2)y)/(dx^(2))+y=0` , iii) `sin^(-1)(dy)/(dx)=x+y`, iv) `log_(e)(dy)/(dx)=ax+by` v) `y(d^(2)y)/(dx^(2))+x((dy)/(dx))^(2)-4y(dy)/(dx)=0` |
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Answer» i) `(dy)/(dx) +y=1/(dy)/(dx)` `therefore ((dy)/(dx))^(2)+y(dy)/(dx)=1` Hence order is 1 and degree is 2. ii) `e^(d^(3)y)/(dx^(3))-x(d^(2)y)/(dx^(2))+y=0` Clearly order is 3, but degree is not defined as it cannot be expressed as a polynomial equation in derivatives. iii) `sin^(-1)(dy)/(dx)=x+y` `therefore (dy)/(dx)=sin(x+y)` Hence order is 1 and degree is 1. iv) `log_(e)(dy)/(dx)=ax+by` `therefore (dy)/(dx)=e^(ax+by)` Hence order is one and degree is also 1. v) `y(d^(2)y)/(dx^(2))+x((dy)/(dx))^(2)-4y(dy)/(dx)=0` Clearly, order is 2 and degree is 1. |
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| 129. |
For certain curve `y=f(x)` satisfying `(d^(2)y)/(dx^(2))=6x-4, f(x)` has local minimum value 5 when `x=1` Global maximum value of `y=f(x)` for `x in [0,2]` isA. 5B. 7C. 8D. 9 |
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Answer» Correct Answer - B Integrating `(d^(2)y)/(dx^(2))=6x-4`, we get `(dy)/(dx) = 3x^(2)-4x+A` When `x=1, (dy)/(dx)=0`, so that `A=1`,Hence, `(dy)/(dx) = 3x^(2)-4x+1` Integrating, we get `y=x^(3)-2x^(2)+x+5`. From equation (1), we get the cricitical points `x=1//3, x=1`. At the critical point `x=1/3, (d^(2)y)/(dx^(2))` is negative Therefore, at `x=1//3, y` has a local maximum. At `x=1, (d^(2)y)/(dx^(2))` is positive. Therefore, at `x=1, y` has a local minimum. Also, `f(1) =5, f(1/3)=139/7, f(0) =5, f(2)=7` Hence, the global maximum value =7 and the global minimum value =5 |
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| 130. |
For certain curves y= f(x) satisfying`[d^2y]/[dx^2]= 6x-4`, f(x) has local minimum value 5 whenx=1.9. Number of critical point for y=f(x) for x € [0,2](a) 0 (b)1. c).2 d) 310. Global minimum value y = f(x) for x € [0,2] is(a)5 (b)7 (c)8 d) 911 Global maximum value of y = f(x) for x € [0,2] is(a) 5 (b) 7 (c) 8 (d) 9 |
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Answer» Correct Answer - C Integrating `(d^(2)y)/(dx^(2))=6x-4`, we get `(dy)/(dx) = 3x^(2)-4x+A` When `x=1, (dy)/(dx)=0`, so that `A=1`,Hence, `(dy)/(dx) = 3x^(2)-4x+1` Integrating, we get `y=x^(3)-2x^(2)+x+5`. From equation (1), we get the cricitical points `x=1//3, x=1`. At the critical point `x=1/3, (d^(2)y)/(dx^(2))` is negative Therefore, at `x=1//3, y` has a local maximum. At `x=1, (d^(2)y)/(dx^(2))` is positive. Therefore, at `x=1, y` has a local minimum. Also, `f(1) =5, f(1/3)=139/7, f(0) =5, f(2)=7` Hence, the global maximum value =7 and the global minimum value =5 |
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| 131. |
For certain curves y= f(x) satisfying`[d^2y]/[dx^2]= 6x-4`, f(x) has local minimum value 5 whenx=1.9. Number of critical point for y=f(x) for x € [0,2](a) 0 (b)1. c).2 d) 310. Global minimum value y = f(x) for x € [0,2] is(a)5 (b)7 (c)8 d) 911 Global maximum value of y = f(x) for x € [0,2] is(a) 5 (b) 7 (c) 8 (d) 9A. 5B. 7C. 8D. 9 |
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Answer» Correct Answer - A Integrating `(d^(2)y)/(dx^(2))=6x-4`, we get `(dy)/(dx) = 3x^(2)-4x+A` When `x=1, (dy)/(dx)=0`, so that `A=1`,Hence, `(dy)/(dx) = 3x^(2)-4x+1` Integrating, we get `y=x^(3)-2x^(2)+x+5`. From equation (1), we get the cricitical points `x=1//3, x=1`. At the critical point `x=1/3, (d^(2)y)/(dx^(2))` is negative Therefore, at `x=1//3, y` has a local maximum. At `x=1, (d^(2)y)/(dx^(2))` is positive. Therefore, at `x=1, y` has a local minimum. Also, `f(1) =5, f(1/3)=139/7, f(0) =5, f(2)=7` Hence, the global maximum value =7 and the global minimum value =5 |
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| 132. |
The integrating factor of `(1+y^(2)) dx = (tan^(-1)y-x) dy` is -A. `tan^(-1)y`B. `e^(tan^(-1))y`C. `(1)/(1+y^(2))`D. `(1)/(x(1+y^(2)))` |
| Answer» Correct Answer - B | |
| 133. |
The integrating factor of differential equation `cos x (dy)/(dx)+y sin x =1` isA. cos xB. tan xC. sec xD. sin x |
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Answer» Given, that `cosx (dy)/(dx)+y sin x=1` `Rightarrow (dy)/(dx)+y tan x=sec x` Here, P=tan x Q=sec x `IF=e^(intPdx)=e^(inttanxdx)=e^(log secx)` `therefore =sec x` |
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| 134. |
The solution of `(dy)/(dx)-y=1, y(0)=1` is given byA. `xy=-e^(x)`B. `xy=-e^(-x)`C. `xy=-1`D. `y=2e^(x)-1` |
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Answer» Given that, `" "(dy)/(dx)-y=1` `rArr" "(dy)/(dx)=1+y` `rArr" "(dy)/(1+y)=dx` On integrating both sides, we get `" "log(1+y)=x+C" "`...(i) when x=0 and y=1, then `" "log2=0+c` `rArr" "C=log2` The required solution is `" "log(1+y)=x+log2` `rArr" "log((1+y)/(2))=x` `rArr" "(1+y)/(2)=e^(x)` `rArr" "1+y=2e^(x)` k `rArr" "y=2e^(x)-1` |
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| 135. |
The solution of differential equation `cos x (dy)/(dx)+y sin x =1`A. `tanx+tany=k`B. `tanx-tany=k`C. `(tanx)/(tany)=k`D. `tanx.tany=k` |
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Answer» Given that, `" "tanysec^(2)xdx+tanxsec^(2)ydy=0` `rArr" "tansec^(2)xdx=-tanxsec^(2)ydy` `rArr" "(sec^(2)x)/(tanx)dx=(-sec^(2)y)/(tany)dy" "`...(i) On integrating both sides, we have `" "int(sec^(2)x)/(tanx)dx=-int(sec^(2)y)/(tany)dy` Put `tanx=t` in LHS integral, we get `" "sec^(2)xdx=dtrArrsec^(2)xdx=dt` and `" "tany=u` in RHS integral, we get `" "sec^(2)ydy=du` On substituting these values in Eq. (i), we get `" "int(dt)/(t)=-int(du)/(u)` `" "logt=-logu+logk` `rArr" "log(t*u)=logk` `rArr" "log(tanxtany)=logk ` `rArr" "tanxtany=k` |
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| 136. |
The integrating factor of `(xdy)/(dx)-y=x^(4)-3x "is"`A. xB. log xC. `(1)/(x)`D. `-x` |
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Answer» Given that `x(dy)/(dx)-y=x^(4)-3x` `Rightarrow (dy)/(dx)-(y)/(x)=x^(3)-3` `"Here" P=-(1)/(x),Q=x^(3)-3` `IF=e^(intPdx)=e^(int(1)/(x)dx)=e^(-log x)` `=(1)/(x)` |
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| 137. |
Family of curves `y=Ax+A^(3)` is represented by the differential equation of degreeA. 3B. 2C. 1D. none of these |
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Answer» Correct Answer - A We have, `y=Ax+A^(3)rArr (dy)/(dx)=A` On eliminating A, we get `y=x(dy)/(dx)+((dy)/(dx))^(3)` Clearly, it is a differential equaiton of degree 3. |
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| 138. |
The family `y=Ax+A^(3)` of curves is represents by differential equation of degreeA. 1B. 2C. 3D. 4 |
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Answer» Given that, `y=Ax+A^(3)` `Rightarrow (dy)/(dx)=A` [we can differential above equatoin only once because it has only one arbitrary constant] `therefore "Degree=1"` |
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| 139. |
The degree of the differential equation corresponding to the family of curves `y=a(x+a)^(2)`, where a is an arbitrary constant isA. 1B. 2C. 3D. 4 |
| Answer» Correct Answer - C | |
| 140. |
The family `y=Ax+A^(3)` of curves is represents by differential equation of degreeA. 3B. 2C. 1D. not defined |
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Answer» Given family of curves is `y=Ax+A^(3)` `Rightarrow (dy)/(dx)=A` Replacing A by `(dy)/(dx)` in Eq.(i) we get `y=x(dy)/(dx)+((dy)/(dx))^(3)` `therefore "Order"=1` |
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| 141. |
`x(1+y^(2))dx+y(1+x^(2))dy=0` at `(0, 0)`A. `(1+x^(2))(1+y^(2))=0`B. `(1+x^(2))(1+y^(2))=1`C. `x^(2)+y^(2)=1`D. `(1-x^(2))(1-y^(2))=1` |
| Answer» Correct Answer - B | |
| 142. |
The curve for which the slope of the tangent at any point is equal to the ration of the abcissa to the cordinates of the point isA. an ellipseB. parabolaC. circleD. rectangular hyperbola |
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Answer» Slope of tangent to the curve =`(dy)/(dx)` and ratio of abscissa to the ordinate =`(x)/(y)` According to the queston, `(dy)/(dx)=(x)/(y)` `yd y=xdx` On integrating both sides, we get `(y^(2))/(2)=(x^(2))/(2)+C` `Rightarrow (y^(2))/(2)-(x^(2))/(2)=C Rightarrow y^(2)-x^(2)=2C` Which is an equation of reactangular hyperbola. |
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| 143. |
`(dy)/(dx)=2e^(x)y^(3)`, when `x=0, y=(1)/(2)`A. `4e^(-x)+y^(2)=8`B. `4e^(x)+y^(2)=8`C. `4e^(x)+y^(-2)=8`D. `4e^(y)-x^(-2)=8` |
| Answer» Correct Answer - C | |
| 144. |
The general solution of `(dy)/(dx)=2xe^(x^(2)-y)` isA. `e^(x^(2)-y)=C`B. `e^(-y)+e^(x^(2))=C`C. `e^(y)+e^(x^(2))+C`D. `e^(x^(2)+y)=C` |
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Answer» Given that, `" "(dy)/(dx)=2xe^(x^(2)-y)=2xe^(x^(2))*e^(-y)` `rArr" "e^(y)(dy)/(dx)=2xe^(x^(2))` `rArr" "e^(y)dy=2xe^(x^(2))dx` On integrating both sides, we get `" "inte^(y)dy=2intxe^(x^(2))dx` Put `x^(2)=t` in RHS integral, we get `" "2xdx=dt` `" "inte^(y)dy=inte^(t)dt` `rArr" "e^(y)=e^(t)+C` `rArr" "e^(y)=e^(x^(2))+C` |
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| 145. |
`(dy)/(dx)-e^(x)=ye^(x)`, when `x=0, y=1`A. `log(1+y)=e^(x)+log2-1`B. `log(2+y)=e^(x)+log2`C. `logy=1+e^(x)log2`D. `log(1+x)+e^(y)+log2=1` |
| Answer» Correct Answer - A | |
| 146. |
STATEMENT-1 : Solution of the differential equation `xdy-y dx = y dy` is `ye^(x//y) = c`. and STATEMENT-2 : Given differential equation can be re-written as `d((x)/(y)) = -(dy)/(y)`.A. Statement-1 is true, Statement-2 is true, Statement-2 is a correct explanation for Statement-1.B. Statement-1 is true, Statement-2 is true, Statement-2 is NOT a correct explanation for Statement-1.C. Statement-1 is True, Statement-2 is FalseD. Statement-1 is False, Statement-2 is True |
| Answer» Correct Answer - A | |
| 147. |
If `(dy)/(dx)-y log_(e) 2 = 2^(sin x)(cos x -1) log_(e) 2`, then y =A. `2^(sin x) + c2^(x)`B. `2^(cos x)+c2^(x)`C. `2^(sin x) + c2^(-x)`D. `2^(cos x) + c2^(-x)` |
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Answer» Correct Answer - A `(dy)/(dx)-y log_(e) 2 = 2^(sin x) (cos x - 1) log_(e) 2` This is linear differential equation I.F. `=e^(-log_(e) 2int dx) = e^(-x log_(e) 2) = 2^(-x)` Solution is `y 2^(-x) = int 2^(-x) 2^(sin x) (cos x - 1) log_(e) 2 dx` put `sin x - x = t rArr (cos x -1) dx = dt` `therefore" "y 2^(-x) = log_(e) 2 int 2^(t) dt` `therefore" "y 2^(-x) = 2^(t) + c` `therefore" "y = 2^(x+t) + c 2^(x)` `therefore" "y = 2^(sin x) + c2^(x)` |
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| 148. |
If `y_(1)(x)` is a solution of the differential equation `(dy)/(dx)-f(x)y = 0`, then a solution of the differential equation `(dy)/(dx) + f(x) y = r(x)` isA. `y = (1)/(y_(1)(x))int r(x) y_(1)(x) dx +(c)/(y_(1)(x))`B. `y = y_(1)(x) int (r(x))/(y_(1)(x))dx + c`C. `y = int r(x) y_(1)(x) dx + c`D. None of these |
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Answer» Correct Answer - A `(dy)/(dx)-f(x)*y=0` `therefore" "(dy)/(y) = f(x) dx` `therefore" "In y = int f(x) dx` `therefore" "y_(1)(x) = e^(int f(x) dx)` Then for given equation, I.F `= e^(int f(x) dx)` Hence, solution is `y*y_(1)(x) = int r(x)*y_(1)(x) dx + c` `y = (1)/(y_(1)(x)) int r(x) * y_(1)(x)dx + (c)/(y_(1)(x))` |
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| 149. |
If `ye^(y) dx = (y^(3) + 2xe^(y))dy, y(0) = 1`, then the value of x when y = 0 isA. `-1`B. 0C. 1D. 2 |
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Answer» Correct Answer - B `ye^(y) dx = (y^(3)+2xe^(y))dy` `rArr" "(dx)/(dy)=(y^(2))/(e^(y))+(2x)/(y)` `rArr" "(dx)/(dy)-(2)/(y)x=(y^(2))/(e^(y))` I.F. `= e^(-2 Iny) = (1)/(y^(2))` General solution `x (1)/(y^(2))= int (y^(2))/(e^(y))-(1)/(y^(2)) dy + c` `(x)/(y^(2))=e^(-y) + c = -e^(-y) + c` `x = 0, y = 1 rArr c = e^(-1)` `rArr" "x = -y^(2)(e^(-y) - e^(-1))` If y = 0, then x = 0 |
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| 150. |
Form differential equation for `(x-a)^(2)+y^(2)=a^(2)` A)`(x+yy_(1))^(2)=y^(2)` B)`(xy_(1)+y)^(2)=y^(2)` C)`y^(2)=x^(2)+2xyy_(1)` D)`x^(2)=y^(2)+2xyy_(1)`A. `(x+yy_(1))^(2)=y^(2)`B. `(xy_(1)+y)^(2)=y^(2)`C. `y^(2)=x^(2)+2xyy_(1)`D. `x^(2)=y^(2)+2xyy_(1)` |
| Answer» Correct Answer - C | |